Chapter 22, Problem 22.62QA

Interpretation Introduction

To write:

The equilibrium constant expression for the reaction between ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)$ and $Ca{F}_2$ using $K_{sp}$ values for the two given reactions and calculate $K$ for the reaction between ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)$ and $Ca{F}_2$.

Answer to Problem 22.62QA

Solution:

The equilibrium constant expression for the reaction between ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)$ and $Ca{F}_2$ is

$K= \frac{{\left[P{O}_4^{3-}\right]}^3\left[O{H}^{-}\right]}{{\left[F^{-}\right]}^{10}}$

The equilibrium constant value for the reaction between ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ and fluoride ions that form $Ca{F}_2$ is $2.5\times {10}^{-7}$.

Explanation of Solution

1) Concept:

We are asked to find the reaction between ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)$ and $Ca{F}_2$. We are given the solubility expressions for dissolution of ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)$ and $Ca{F}_2$ and using the $K_{sp}$ values for the two reactions, we can determine the $K$ value for the reaction between ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)$ and $Ca{F}_2$.

K value of a reaction running in reverse is the reciprocal of K for the forward reaction.If the original chemical equation describing an equilibrium is multiplied bya factor n, the value of K of the new equilibrium constant expression is the value of the original K raised to the nth power.If an overall chemical reaction is the sum of two or more other reactions, the overall value of K is the product of the K values of the other reactions. .

For this, we should reverse the second equation and multiply it by a coefficient 5 to get the desired coefficient for the reaction between ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)$ and $Ca{F}_2$ and then, add this modified second reaction to the first given reaction to get the overall reaction between ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)$ and $Ca{F}_2$.

2) Given:

Solubility expressions for ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ and $Ca{F}_2$ are given as

${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)\rightleftharpoons 5 {Ca}^{2+}\left(aq\right)\rightleftharpoons 3{PO}_4^{3-}\left(aq\right)+{OH}^{-}\left(aq\right)$; $K_{sp}=2.3\times {10}^{-59}$

$Ca{F}_2\left(s\right)\rightleftharpoons {Ca}^{2+}\left(aq\right)+F^{-}\left(aq\right)$; $K_{sp}=3.9\times {10}^{-11}$

The overall reaction between ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ and $Ca{F}_2$ is:

${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)+10 F^{-}\left(aq\right)\rightleftharpoons 5 Ca{F}_2\left(s\right)+3{PO}_4^{3-}\left(aq\right)+{OH}^{-}\left(aq\right)$

3) Calculations:

The given two reactions are

1) ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)\rightleftharpoons 5 {Ca}^{2+}\left(aq\right)\rightleftharpoons 3{PO}_4^{3-}\left(aq\right)+{OH}^{-}\left(aq\right)$$K_1=2.3\times {10}^{-59}$

2) $Ca{F}_2\left(s\right)\rightleftharpoons {Ca}^{2+}\left(aq\right)+2F^{-}\left(aq\right)$${ K}_2=3.9\times {10}^{-11}$

Overall reaction for which we need to find the value of K is

${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)+10 F^{-}\left(aq\right)\rightleftharpoons 5 Ca{F}_2\left(s\right)+3{PO}_4^{3-}\left(aq\right)+{OH}^{-}\left(aq\right)$

The equilibrium constant expression for the overall reaction is written as

$K= \frac{{\left[P{O}_4^{3-}\right]}^3\left[O{H}^{-}\right]}{{\left[F^{-}\right]}^{10}}$

Overall reaction shows that, there are 10 fluoride ions on the reactant side, so we need to first reverse the eq. (2),

${Ca}^{2+}\left(aq\right)+{2F}^{-}\left(aq\right)\rightleftharpoons Ca{F}_2\left(s\right)$

Therefore, the new equilibrium constant $\left(K_3\right)$ will be the reciprocal of $K_2$ i.e.

$K_3=\frac{1}{K_2}=\frac{1}{3.9\times {10}^{-11}}=2.564\times {10}^{10}$

Hence,

3) ${Ca}^{2+}\left(aq\right)+2F^{-}\left(aq\right)\rightleftharpoons Ca{F}_2\left(s\right); K_3=2.564\times {10}^{10}$

Now, multiply equation (3) by 5 to get 10 fluoride ions as in overall reaction, and the equilibrium constant increases by the power of 5.

${5Ca}^{2+}\left(aq\right)+10F^{-}\left(aq\right)\rightleftharpoons 5Ca{F}_2\left(s\right)$

$K_4={\left(K_3\right)}^5= {\left(2.56410\times {10}^{10}\right)}^5=1.10835\times {10}^{52}$

i.e.

4) ${5Ca}^{2+}\left(aq\right)+10F^{-}\left(aq\right)\rightleftharpoons 5Ca{F}_2\left(s\right); { K}_4=1.10835\times {10}^{52}$

Now, add equation (1) and (4) and the new equilibrium constant is the multiplication of $K_1$ and $K_4$ is

${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)\rightleftharpoons 5 {Ca}^{2+}\left(aq\right)\rightleftharpoons 3{PO}_4^{3-}\left(aq\right)+{OH}^{-}\left(aq\right)$ $K_1=2.3\times {10}^{-59}$

${5Ca}^{2+}\left(aq\right)+10F^{-}\left(aq\right)\rightleftharpoons 5Ca{F}_2\left(s\right)$ $K_4=1.10835\times {10}^{52}$

${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)\left(s\right)+10 F^{-}\left(aq\right)\rightleftharpoons 5 Ca{F}_2\left(s\right)+3{PO}_4^{3-}\left(aq\right)+{OH}^{-}\left(aq\right)$ $K=K_1\times K_4$

$K=2.3\times {10}^{-59}\times 1.10835\times {10}^{52}$

$K=2.5492\times {10}^{-7}$

$K=2.5\times {10}^{-7}$

Therefore, $K$ for the reaction between ${Ca}_5{\left(P{O}_4\right)}_3\left(OH\right)$ and fluoride ions that forms $Ca{F}_2$ is $2.5\times {10}^{-7}$.

Conclusion:

We have manipulated the individual given reactions to get the equilibrium constant value for the overall reaction which is obtained by combining them.