M5 EQUIL 3-4

IQ3  How can the position of equilibrium be described, and what does the equilibrium constant represent?

IQ3. SYLLABUS POINTS 

5.3.1 The equilibrium expression and constant Keq.

5.3.2 Calculating Keq. 

5.3.3 Temperature and Keq. 

5.3.4 Investigating Keq. 

5.3.5 Keq and dissociation reactions. 

Consider the following equilibrium general reaction

                        aA(aq)    +    bB(aq)        ⇌    cC(aq)    +     dD(aq)

At equilibrium, the ratio of the multiplication (arithmetic product) of the concentrations of the substances on the right hand side of the equation divided by the multiplication (arithmetic product) of the concentrations of substances on the left hand side is a numerical constant – called the equilibrium constant. 

We have an acronym which can help us to remember this relationship: PoRK!

Products over Reactants = K

Where there is more than one molecule (or mole) of a substance involved in the reaction, we raise the value of the concentration of that substance to a power equal to the number of molecules (or moles) indicated. For the reaction above, the equilibrium expression would be as follows:

                        [C]c x [D]d                  

            Keq  =     _________                           

                        [A]a x [B]b                              the value of K is constant for a given temperature. 

To determine whether a particular reaction has reached equilibrium, we can calculate the concentration quotient Q, and compare it to Keq. Consider Q to be like a test for equilibrium. It is the same expression, only the values of the concentrations are different.

An expression of Q for the reaction:

            2CrO42−(aq)  +  2H+(aq)  ⇌  Cr2O72−(aq)  +  H2O(l)

would be:

                                                [Cr2O72−] x [H2O]

                                    Q  =     ___________________

                                                  [CrO42−]2 x [H+]2

NB: We can eliminate water from this calculation as it is a liquid and has a constant concentration. So the adjusted equation would be:

                                                        [Cr2O72−]

                                    Q  =     _____________________

                                                  [CrO42−]2 x [H+]2

If the value of Q = the value of Keq then the reaction has reached equilibrium.

If the value of Q < the value of Keq the concentrations of the reactants are higher than they would be at equilibrium, the reaction will continue to go in the forward direction until it reaches equilibrium.

If the value of Q > the value of Keq the concentrations of the products are higher than they would be at equilibrium, the reaction will proceed in the reverse direction to reach equilibrium.

Le Châtelier’s Principle and the Equilibrium Constant

When you apply Le Chatelier’s principle to an equilibrium problem involving a change in concentration, you might assume that Keq must change. This seems logical since we talk about “shifting” the equilibrium in one direction or the other. However, Keq is a constant, for a given equilibrium at a given temperature, so it does not change. Here is an example of how this works. Consider the simplified equilibrium below:

A⇄B

Let’s say we have a 1.0 litre container. At equilibrium the following amounts are measured.

A = 0.50 mol,  B = 1.0 mol

The value of Keq is given by:

Keq=  [B] / [A]  =  1.0 M / 0.50 M  

= 2.0

Disturb the equilibrium by adding 0.50 mole of A to the mixture. At that instant, before the system responds to the change, the amounts are 1.0 mol of A and 1.0 mol of B in 1 litre (so their concentrations are now 1.0M and 1.0M).  

With an increase in concentration of A, the equilibrium will shift to the right, using up the added A, but by how much?

Use X to represent the change in concentrations as the reaction proceeds. Since the mole ratio of A:B is 1:1, as [A] decreases by X, [B] increases by X. 

We set up an analysis called ICE, which stands for Initial, Change, and Equilibrium. The values in the table represent molar concentrations.

                            A                     B

             ––––––––––––––––––––

Initial                 1.0                1.0

Change             −X                   +X

Equilibrium    1.0−X           1.0+X

At the new equilibrium position, we can solve for X.

Keq=  [B] / [A]

We know Keq = 2.0, so

2.0  =  [B] / [A]

Substituting in the amounts, 

2.0  = 1.0+X / 1.0−X

Solving for X:

2.0(1.0−X) = 1.0+X

2.0−2.0X  =1.0+X

3.0X = 1.0

X=0.33

This value for X is now plugged back into the E (Equilibrium) line of the RICE table to calculate the final concentrations of A and B after the reaction.

[A]=1.0−X=0.67 M              [B]=1.0+X=1.33 M

The value of Keq has been maintained, since 1.33 / 0.67=2.0. 

A change in concentration of one of the substances in equilibrium causes a shift in the equilibrium position, but the value of the equilibrium constant does not change.

VIEW Videos 

• EAR#11 https://www.youtube.com/watch?v=HmGn6IF7UoU&index=13&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz  [7.26 mins]

• Equilibrium Constant https://www.youtube.com/watch?v=xfGlEXWDRZE  [6.14 mins]

• Equilibrium https://www.youtube.com/watch?v=cHAjhM3y3ds  [12.23 mins]

• Reaction Quotient https://www.youtube.com/watch?v=4FRWEuZIP0o&index=8&list=PLIYT0Z0pGeLyaKnFtf3JPJa3lkoW25sRw  [6.36 mins]

While we cannot compare one K value directly with another (actually we can if they have equal units for K), the magnitude of the K value for a given reaction tells us something about the equilibrium position for that reaction.

When K is large:      (> 103) 

The equilibrium lies well to the right, the concentration of products will be very high compared to the concentration of the reactants.

eg.       Zn2+     +          4CN-   ⇌        Zn(CN)42-                   K  =  1017 mol-4 L-4 (at 298K)

When K » 1:              (0.1 - 10) 

The equilibrium lies in the middle, neither favouring reactants nor products, both will be in similar concentrations.

 eg.      N2O4(g)                                ⇌        2NO2(g)                                   K = 0.48 mol L-1 (at 373K)

When K is small:     (< 10-3) 

The equilibrium position lies well to the left, the concentration of reactants will be very high compared to the concentration of the products.

eg.       2SO3(g)             ⇌        2SO2(g)      +        O2(g)                 K = 10-12 mol L-1 (at 500 K)

To answer Keq questions:

1.     Write out a balanced chemical equation.

2.     Write out the general expression for the Keq of the system.

3.     Write out a table (remember RICE: Ratio, Initial, Change, Equilibrium/Final)

4.   Fill in the table, find all values and solve for the desired quantity. 

How do we deal with chemical reactions where one or more gases are in equilibria with solids and/or liquids?

Gases in equilibrium with solids or liquids.

Example 1.               CaCO3(s)          ⇌        CaO(s)              +          CO2(g)

In this case both [CaCO3] and [CaO] are constants, hence K simply = [CO2].  At a given temperature, the concentration of CO2 in equilibrium with the solids is a measure of the position of the equilibrium.

Example  2.               H2O(l)              ⇌        H2O(g)

As the concentration of a pure liquid is a constant, the value of K is modified to give the expression K = [H2O(g)]. This is a measure of the saturated vapour pressure of water at the given temperature.

Extension       2NO(g) + O2(g)   ⇌  2NO2(g)

In this case of homogeneous equilibria, we calculate Keq using concentrations, or we may be able to calculate KP, based on the partial pressures of each of the three gas species, assuming that PT = sum of the pressures of each component in the mixture. To do this, we need to know the mole fraction for each gaseous species as well as the partial pressure exerted by each gas. For the example above, to calculate the mole fractions, let’s assume we have 2 + 1 + 2 = 5 moles of gas in total at equilibrium. Then 2 is NO, 1 is O2 and 2 is NO2. If we assume 100kPa of pressure, then the partial pressures will be in the same ratios as the mole fractions multiplied by the total pressure (assumption is the partial pressures exerted by each component gas add up to the total pressure in the container). In this case  2/5 x 100 = 40kPa for NO,  1/5 x 100 = 20kPa for O2 and  2/5 x 100 = 40kPa for NO2. 

So KP = P2(NO2) / P2(NO)xP(O2) or 402 / 402 x 20  = 0.05. 

There is no clear indication about whether you need to look at partial pressure calculations, so investigate this further only if you are interested.

VIEW Videos:

•  EAR#12 https://www.youtube.com/watch?v=fl85PSWTiRI&index=12&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz [7.16 mins] 

• Equilibrium Equations: Crash Course Chemistry 29 https://www.youtube.com/watch?v=DP-vWN1yXrY [9.28 mins]

• Professor Dave: Equilibrium Calculations https://www.youtube.com/watch?v=1GiZzCzmO5Q  [6.47 mins]

WORKED EXAMPLES:

• For Example 3, where they write atmospheres, read that as moles.

ANSWERS:

Q1

• (a) [H2O] is included as it is a gas,  but the concentration of the sulfur isn't as it is a solid. 

• (b) The best responses included full workings to demonstrate how the value of each of the equilibrium concentrations was determined. 

• (c) Increasing [SO2], increasing pressure, decreasing volume, decreasing temperature. The equilibrium position is not affected by the removing the solid sulfur. 

Q2

Q3  (c)(i) In the better responses, candidates recognised that the equilibrium expression had to be used to perform a calculation to determine if the system was at equilibrium. Weaker responses did not recognise that the container was 5.0 L, leading to an incorrect calculation of Q. Poorer responses made use of addition instead of multiplication of concentrations in the equilibrium expression or attempted to answer the question without referring to the data provided. 

(ii) In the best responses, candidates provided explanations that made use of Le Chatelier’s principle. Better responses correctly explained the influence of pressure on the system and correctly related the endothermic reaction to the absorption of heat after increasing the temperature of the system. Weaker responses correctly identified the conditions required, but neglected to explain these conditions in terms of Le Chatelier’s principle, or simply stated Le Chatelier’s principle without relating it to the system involved. 

"qualitatively analyse", means be able to say if it will increase/decrease etc. No calculation / numbers (quantitative) required. 

Equilibrium is reached when the forward and reverse reaction rates are equal. These rates are determined by the frequency of successful collisions, which is determined by 

·        concentration 

·        temperature

·        activation energy

·        how often molecules collide with the correct orientation (some catalysts work by “holding” the reactants to provide the correct orientation, speeding up the reaction)

Changes in concentration, pressure and volume may alter the equilibrium position, but do not change the ratio of products to reactants, therefore they do not change the value of K, the equilibrium constant. 

Change in temperature does alter the equilibrium constant.

An increase in temperature will increase kforward and kreverse , but by different amounts because their activation energies are rarely equal. This changes K, which is the ratio of the two rate constants.

Temperature is the only factor that changes the equilibrium constant for a given reaction. 

A rise in temperature will increase the value of K for endothermic reactions and decrease it for exothermic reactions. 

A decrease in temperature will decrease the value of K for endothermic reactions and increase it for exothermic reactions. 

(Changing other factors may cause a shift in the equilibrium position, defined by the reaction quotient Q (the position of the reaction if not at equilibrium), but after the system has responded and equilibrium is re-established, there will have been no change to the K value.)

Consider the following equilibrium:

N2(g)  +  3H2(g)  ⇌  2NH3(g)                 ΔH = -ve

As this is an exothermic reaction, we could regard energy as a product of this reaction.

N2(g)  +  3H2(g)  ⇌  2NH3(g)  +  energy

For this (exothermic) reaction, an increase in temperature will increase the reverse reaction, producing more reactants (increasing the size of the denominator), and so lowering the value of K. As the temperature increases, the equilibrium shifts to the left. This increases the concentration of the reactants (denominator) and so decreases  the value of K.

Consider the exothermic equation:          2SO2(g)  +  O2(g)  ⇌  2SO3(g)  +  heat

This is an important reaction in the manufacture of sulfuric acid (from SO3).  Highest yields of SO3 occur at low temperatures, however very low temperatures reduce reaction rates and thus overall yields. Once again compromise temperatures (450oC) and catalysts are used. Excess oxygen is also used to drive the equation towards the right.

Consider the endothermic equation:        N2O4(g)  +  heat  ⇌  2NO2(g)

As the temperature increases, the equilibrium shifts to the right. This increases the concentration of the product (numerator) and so increases  the value of K.

ATAR Notes Key Points:

•   Only changes in temperature will change the value of Keq. 

•   Increasing the temperature in an exothermic reaction will decrease Keq.

•   Increasing the temperature in an endothermic reaction will increase Keq. (Silove, 2018, p. 20)

Did I mention temperature is the only factor that changes the equilibrium constant for a given reaction. ????

Why not concentration?

Consider the general reaction: A  +   B    ⇌   C   +   D

The equilibrium constant, Kc for this reaction looks like this:

  Keq  =  [C]  x  [D]   /    [A]  x  [B]

If you increase the concentration of A,  which will move the position of the equilibrium to the right (and so increase the amount of C and D), why hasn't the equilibrium constant changed?

If you increase the concentration of A, the bottom of the Kc expression gets larger. That would change the value of Kc. But an increase in A causes it to react more with B, reducing that, and produces more C and D. The top of the expression gets larger. A new balance is reached and the value of the equilibrium constant expression reverts to what it was before.

The position of equilibrium moves but not the value of Keq. There is a constant value for the equilibrium constant.

A similar thing happens for pressure, so it also does not change the value of Keq.

VIEW Videos:

• EAR#13 https://www.youtube.com/watch?v=udmeoshSIOQ&index=11&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz [7.39 mins]

• Temperature and Keq  https://www.youtube.com/watch?v=Pa9bgjp81EM&index=9&list=PLIYT0Z0pGeLyaKnFtf3JPJa3lkoW25sRw  [8.11 mins]

ANSWERS:

1. 

6. b i)

6 b ii) Sample answer: Since the reaction as written is endothermic, cooling the reaction vessel will drive the reaction to the left, reducing the concentration of purple I2. Hence, contents of vessel will change to a lighter shade of purple. 

VIEW Videos 

• EAR#14 https://www.youtube.com/watch?v=o7VzsSIl2rw&index=16&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz [7.49 mins]

• Using colourimeter with Vernier calculations https://www.youtube.com/watch?v=xs8T_dKsPjg [6.11 mins]

• Determination of pKa of weak acid using pH meter  https://www.youtube.com/watch?v=y3Oa0L404oM 

When a salt is dissolved in water, the crystal is broken down as water molecules draw off the ions (the ions are hydrated). 

This continues to occur until the solution becomes saturated (no more solid will dissolve). 

This now also represents an equilibrium as the rate of dissociation of solid from the crystal, is balanced by the rate of association of the salt (as ions in solution combine to form the solid crystal.) 

eg.       NaCl(s)                        ⇌        Na+      +            Cl-                               ΔH  =  + ve

The equilibrium constant for saturated solution and solid formation (precipitate) is called the solubility product, Ksp. 

For unsaturated and supersaturated solutions, the system is not at equilibrium. For these the ion product, Qsp, which has the same expression as Ksp, is used. 

We will look at the Ksp in Section 4.

Explore the Virtual Lab

• https://www.learner.org/courses/essential/physicalsci/session3/interactive.html 

Likewise, the negatively charged region of the water molecules interact with the positively charged ions (sodium) drawing them away from the negative ions. 

At SLC, NaCl has a solid, crystalline structure. When NaCl is added to liquid water, it loses its crystalline structure. This is because NaCl dissociates to form Na+ and Cl- ions. 

The NaCl lattice is dissociated by water molecules that pull Na+ and Cl- ions away from the lattice. This can be summarised as follows:

•      Water is a polar molecule that contains a slightly negative end (O δ-) and two slightly positive (H δ+) ends.

•      Due to the electrostatic attraction, the H δ+ end of the molecule is attracted to Cl- ions, while the O δ- end is attracted to the Na+ ions.

•      The electrostatic attraction that exists between these particles disrupts the balance of the ionic bond that exists between Na+ and Cl-  , and  the repulsion the Na+ ions have for each other and the repulsion the Cl-  ions have for each other. 

•      As a result, the ions attach to, and are surrounded by, the water molecules and are broken away from the crystal structure. 

We can refer to the Na+ and Cl- ions in the aqueous solution as ‘hydrated’ because they move freely in the aqueous solution.

Not all ionic substances are soluble in water, because in these compounds, the electrostatic attraction of the ions to water molecules is not strong enough to disrupt the balance of the crystal lattice. 

Ion equilibrium

Once an equilibrium has been established in an ionic solution, we can upset the equilibrium in a number of ways.

•          Increasing the temperature shifts the equilibrium to the right.

•          Adding K+ ions to the solution shifts the equilibrium to the left.

•          Adding Cl- ions to the solution shifts the equilibrium to the left.

Adding more KCl will not affect the equilibrium as the concentration of a solid is unchanging. 

Adding ions already present in a solution, causes the equilibrium to shift in order to use up the excess ions (Le Chatelier’s principle).  It is commonly called the common ion effect.  Adding common ions to a solution shifts the equilibrium, however, adding a different salt with no common ions (eg. NaBr to the equilibrium mentioned above) has no effect on the original equilibrium (the KCl in the above case).

The common ion effect is particularly important when using slightly soluble salts, eg. AgCH3COO or Ca(OH)2.  AgCH3COO has a solubility of 1.04 g / 100 mL of H2O. (5.5 x 10-2 mol.L-1) at 25oC.  The presence of any common ions dramatically lowers its solubility. (eg. in a solution of 0.10 mol.L-1 NaCH3COO, the solubility of AgCH3COO drops to 2.4 x 10-2 mol.L-1).

VIEW Videos 

• EAR#17 https://www.youtube.com/watch?v=lqu48E-u_TE&index=18&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz   [6.46 mins]

• Dissolution of Ionic Compounds https://www.youtube.com/watch?v=S5eSvFiStrg&index=10&list=PLIYT0Z0pGeLyaKnFtf3JPJa3lkoW25sRw  [9.36 mins]

VIEW Videos 

• EAR#18 https://www.youtube.com/watch?v=rtbv7PBV4VM&index=17&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz    [7.22 mins]

• Making poisonous blackbean safe  https://www.youtube.com/watch?v=eVvQnsKuOcE  [4.50 mins]

• Cooking blackbeans  https://www.youtube.com/watch?v=hEOAJM42MA4  [1.56 mins]

VIEW Videos 

• EAR#19 https://www.youtube.com/watch?v=6HdyFfPfnXE&index=21&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz     [7.23 mins]

• Precipitation Reactions https://www.youtube.com/watch?v=aiAFtFqHSqs&index=11&list=PLIYT0Z0pGeLyaKnFtf3JPJa3lkoW25sRw [6.23mins]

A saturated solution is one in which no more solute will dissolve in the solvent. Once a solution is saturated, any further addition of solute will be deposited as a precipitate on the bottom of the solution, or held as a suspension.

However, this is a dynamic process. Even though it looks like there is nothing happening at a macro level, if we could see what is happening at a micro level, we would notice that the rate of ions being added to the solution is the same as the rate at which ions are combining to form a solid precipitate.

We call this a dynamic process as both the forward and reverse processes are occurring at the same rate.

For saturated solutions of soluble ionic solids; 

eg.       PbCl2(s)   ⇌     Pb2+(aq) +  2Cl-(aq)  we would usually write for K expression:

                [Pb2+] x [Cl-]2

        K  =    ——————

However [PbCl2] in the solid phase is not affected by how much of it there is.  Twice as much would occupy twice the volume. We can simplify this by using the solubility product constant  Ksp.  This recognises that the concentration of the solid remains constant, so: 

Ksp =   [Pb2+] x [Cl-]2

Like K, the solubility product will also change with temperature eg. lead chloride has a solubility of: 

• 10g / L     at 293 K 

• 33g / L     at 373 K

In general, the more insoluble the substance, the lower the K value,   BUT   Ksp values can only be compared between different substances if they have the same number of ions per formula unit (and therefore the same units). Consider the value of Ksp for NaCl 

Ksp =   [Na+] x [Cl-]         [units will be mol2 L-2   or  M2] 

compared to the value of Ksp for Al2(SO4)3       

Ksp =   [Al3+] 2   x  [SO42-] 3     [units will be M5] 

VIEW Videos: 

• EAR#20 https://www.youtube.com/watch?v=PKdULkbUdsU&index=20&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz    [7.43 mins]

• Determining Molar Solubility Given Ksp https://youtu.be/QWJW1xzB_EA   [4.25 mins]

• Solubility Product  https://www.youtube.com/watch?v=2QZTPsZKGMU&list=PLIYT0Z0pGeLyaKnFtf3JPJa3lkoW25sRw&index=12  [6.23 mins]

The solubility product can be applied to precipitation reactions. As for previous equilibria, it can be helpful to use a RICE table [mole Ratio,  Initial concentration, Change in concentration, Equilibrium concentration] to assist with some of these types of calculations. 

Example 1:

Silver chloride has a molar solubility of 1.33 x 10-5 mol/L at a specific temperature. (Chan, et al., 2019, p. 124)

Calculate the Ksp for silver chloride at this temperature.

            AgCl(s)   ⇌      Ag+(aq) +          Cl-(aq)  

MR      1                      1                      1

E                          1.33 x 10-5            1.33 x 10-5

Ksp        =           [Ag+]          x      [Cl-]

         =          1.33 x 10-5 x 1.33 x 10-5

         =          1.77 x 10-10    M2

Example 2:

Zinc hydroxide has a Ksp value of 3.8 x 10-17 at 25oC. Calculate the molar solubility of zinc hydroxide at this temperature. (Chan, et al., 2019, p. 125)

          Zn(OH)2(s)      ⇌         Zn2+(aq)             +  2OH-(aq) 

We don't know the concentrations, so use the algebra sign of X (capital is just to distinguish it from x used as a multiplication sign) for the zinc hydroxide that dissolves. The equation shows that one zinc hydroxide produces 1 zinc ion and 2 hydroxide ions: the value of hydroxide will be double that of zinc according to the coefficients in the dissociation equation shown below, so it will be 2X.

            Zn(OH)2(s)      ⇌        Zn2+(aq)             +  2OH-(aq) 

MR      1                                  1                            2

E                                             X                               2X

Ksp                   =          [Zn2+]   x   [OH-] 2

3.8 x 10-17        =          X x (2x)2

3.8 x 10-17             =          4x3

taking the cube root and then dividing by 4

2.12 x 10-6       =          x

The molar solubility of zinc hydroxide in water is  = 2.12 x 10-6 mol/L 

Example 3:

Calcium fluoride has a Ksp value of 3.2 x 10-11 at 25oC. If 1.0 x 10-3 g of calcium fluoride is added to 1L of water at 25oC, predict whether a precipitate will form. (Chan, et al., 2019, p. 126)

Calculating the concentration 

moles: n = m /MM   =  0.0010  / {40.08 + (2x19.00)}   =  1.28 x 10-5 mol

When dissolved:

[CaF2(aq)]  =  moles / V  =   1.28 x 10-5 mol / 1 L  =  1.28 x 10-5 mol/L

            CaF2(s)   ⇌       Ca2+(aq)             +  2F-(aq) 

R             1                      1                               2

E                          =  1.28 x 10-5                 2.56 x 10-5                

Ksp                   =          [Ca2+] x [F-]2

                        =          1.28 x 10-5  x  (2.56 x 10-5)2

                        =          8.40 x 10-15    M3

Since 8.40 x 10-15  is less than 3.2 x 10-11,  there will be no precipitate.

Example 4:

5.0 mL of 0.010 mol.L-1 barium chloride and 20 mL of 0.010 mol.L-1 sodium sulfate are mixed together. Predict whether a precipitate will form, and, if so, identify the precipitate. (Chan, et al., 2019, p. 127)

There are two possible precipitates, barium sulfate and sodium chloride. From solubility rules, we know that sodium chloride is soluble, so we are looking to see if barium sulfate will precipitate.

Dissociation: 

BaCl2(aq)     ⇌    Ba2+(aq)  +     2Cl-(aq)       

Finding the concentration of BaCl2 when it is added to Na2SO4 :    

M1V1  = M2V2  

 0.01  x  0.005   =   M2 x 0.025  (since 5 mL is added to 20 mL = 25 mL  =  0.025 L)

M2 = 0.01 x 0.005 / 0.025  =  0.002 mol/L

Finding the concentration of barium ions in the mixed solution: 

BaCl2(aq)     ⇌    Ba2+(aq)  +     2Cl-(aq)      

1                           1                      2

1 mol BaCl2   dissociates to give 1 mol barium ions, so [Ba2+]  =  0.002 mol/L

Finding the concentration of Na2SO4  when it is added to BaCl2:

M1V1  = M2V2  

0.010  x 0.020  =   M2 x 0.025

M2 = 0.010 x 0.020  / 0.025  = 0.008 mol/L

Finding the concentration of sulfate ions in the mixed solution: 

Na2SO4(aq)   ⇌      2Na+(aq)  +     SO42-(aq)  

1                                  2                   1

1 mol Na2SO4   dissociates to give 1 mol sulfate ions, so [SO42-]  =  0.008 mol/L

   BaCl2(aq)   +  Na2SO4(aq)   ⇌    Ba2+(aq)  +          SO42-(aq)  +        2Na+(aq)  +       2Cl-(aq)  

R      1                      1                      1                            1                        2                      2

The ionic lattice of BaSO4   is a ratio of 1:1 for the two ions. Since there is only 0.002 mol/L of barium, it is a limiting reagent. Only 0.002 of the sulfate ion will react,  the excess will stay in solution. The reacting quantities will be 0.002 mol/L barium and sulfate ions:                               

 Ionic product (Q)     =          [Ba2+] x [SO42-]

                                    =          0.002 x 0.002

                                    =          4.00 x 10-6    M2

Ksp for BaSO4 is 1.08 x 10-10. 

As Q (the ionic product) > Ksp, a precipitate will form.

ATAR Notes Key Point:

To determine if a precipitate forms:

1.     Find the limiting reagent.

2.     Calculate the solubility.

3.   Determine whether there is a precipitate remaining. (Silove, 2018, p. 27) 

VIEW Videos 

• EAR#21 https://www.youtube.com/watch?v=3gd4-4LsrZw&index=19&list=PLeFSFSJ9WqSA3w0P26A6m-NV-gA1Y7Hiz    [9.31 mins]

• Determine Ksp given solubility https://youtu.be/yCKfF4ylEfE   [5.25 mins]

VIEW Videos:

• Comparing Qsp and Ksp to Determine Whether a Precipitate Will Form  https://www.youtube.com/watch?v=K7WlQicX2nE&feature=youtu.be    [5.53 mins]

• Calculations involving the solubility product https://www.youtube.com/watch?v=qafaZsra6po&index=13&list=PLIYT0Z0pGeLyaKnFtf3JPJa3lkoW25sRw  [9.19 mins]