Question

predict the equilibrium concentration of ca²⁺ in the reaction described below (for which kc = 7.1×10⁻⁴ at the reaction temperature) by constructing an ice table, writing the equilibrium constant expression, and solving for the equilibrium concentration. complete parts 1 - 3 before submitting your answer. cacro₄(s) ⇌ ca²⁺(aq) + cro₄²⁻(aq) a solid sample of cacro₄ dissolves in solution. fill in the ice table with the appropriate value for each involved species to determine the concentration of all reactants and products. initial (m) change (m) equilibrium (m) cacro₄(s) - - - ca²⁺(aq) + cro₄²⁻(aq) +

Explanation:

Step1: Write the equilibrium - constant expression

For the reaction $CaCrO_4(s)
ightleftharpoons Ca^{2 + }(aq)+CrO_4^{2 - }(aq)$, the equilibrium - constant expression $K_c$ is given by $K_c=[Ca^{2 + }][CrO_4^{2 - }]$. Since $CaCrO_4$ is a solid, it is not included in the $K_c$ expression.

Step2: Set up the ICE table

Let the initial concentration of $Ca^{2 + }$ and $CrO_4^{2 - }$ be $0$ (assuming no ions present initially before dissolution). Let the change in concentration of $Ca^{2 + }$ and $CrO_4^{2 - }$ due to the dissolution of $CaCrO_4$ be $x$.

	Initial (M)	Change (M)	Equilibrium (M)
$Ca^{2 + }(aq)$	$0$	$x$	$x$
$CrO_4^{2 - }(aq)$	$0$	$x$	$x$

Step3: Solve for $x$

We know that $K_c = 7.1\times10^{-4}$. Substituting into the $K_c$ expression $K_c=[Ca^{2 + }][CrO_4^{2 - }]$, we get $7.1\times10^{-4}=x\times x=x^{2}$. Solving for $x$, we take the square - root of both sides: $x=\sqrt{7.1\times10^{-4}}\approx2.7\times10^{-2}\ M$.

Answer:

The equilibrium concentration of $Ca^{2 + }$ is $2.7\times10^{-2}\ M$.