What is the reaction quotient, Q, for the equilibrium
CuCl(s) $\rightleftharpoons$ Cu⁺(aq) + Cl?(aq)
When 0.3746 L of $4.359 \times 10 ^ { - 4 }$ M Cu⁺ is combined with 0.4326 L of $8.718 \times 10 ^ { - 4 }$ M Cl? in the presence of an excess of CuCl(s) ?

Question

What is the reaction quotient, Q, for the equilibrium
CuCl(s)

\rightleftharpoons

Cu⁺(aq) + Cl?(aq)
When 0.3746 L of

4.359 \times 10 ^ { - 4 }

M Cu⁺ is combined with 0.4326 L of

8.718 \times 10 ^ { - 4 }

M Cl? in the presence of an excess of CuCl(s) ?

Quizplus · Accepted Answer

The reaction quotient is given by the expression:$$Q = \frac{[Cu^+][Cl^-]}{[CuCl(s)]}$$Since CuCl(s) is a solid, its concentration is not included in the expression for Q. Therefore, we can simplify the expression to:$$Q = [Cu^+][Cl^-]$$We can calculate the concentrations of Cu+ and Cl- using the given volumes and concentrations:$$[Cu^+] = \frac{0.3746 L \times 4.359 \times 10^{-4} M}{0.3746 L + 0.4326 L} = 2.179 \times 10^{-4} M$$$$[Cl^-] = \frac{0.4326 L \times 8.718 \times 10^{-4} M}{0.3746 L + 0.4326 L} = 4.359 \times 10^{-4} M$$Substituting these concentrations into the expression for Q, we get:$$Q = (2.179 \times 10^{-4} M)(4.359 \times 10^{-4} M) = 9.451 \times 10^{-8}$$Therefore, the reaction quotient is $9.451 \times 10 ^ { - 8 }$.

What Is the Reaction Quotient, Q, for the Equilibrium ⇌\rightleftharpoons⇌

What Is the Reaction Quotient, Q, for the Equilibrium
$\rightleftharpoons$