Recently I've been trying to answer the question 10-13 (c) of 9th edition of Fundamentals of Analytical Chemistry of Skoog et al. Comparing my answers with the student manual , I realized that my numbers didn't agree to the amount of $\mu$ of solution calculated by the book. Here's the full question:
Use activities to calculate the molar solubility of $Zn(OH)_2$ in the solution that results when you mix 40.0 mL of 0.250M $KOH$ with 60.0 mL of 0.0250M $ZnCl_2$.
(Available information : $K_{sp}$ $Zn(OH)_2$ = $ 3\times 10^{-16}$ , $\alpha_{Zn^{+2}}$ = 0.6 nm , $\alpha_{OH^{-}}$=0.35 nm , $-\log\gamma_x=\frac{0.51{Z_x}^2 \sqrt{\mu}}{1+3.3\alpha_x \sqrt{\mu}}$)
I'm not looking for the full answer. I'm okay with the whole provided answer except with $\mu$. To be brief the final solution should contain $[K^+]=0.1M$ , $[OH^-]=0.07M $ , $[Cl^-]=0.03M $ , $[Zn^{+2}] \approx 0 M$. Then the manual says : $\mu = {1\over 2 }{(0.1 \times 1^2 +0.07 \times 1^2 + (2) \times 0.03 \times 1^2) }$ which means $\mu = 0.115$ . From that, the solubility is found to be 2.8 $\times$ $10^{-13}$ M
From my point of view: $\mu={1\over 2 }{(0.1 \times 1^2 +0.07 \times 1^2 + 0.03 \times 1^2) }$
Therefore $\mu=0.1M$ and finally Solubility = 2.6302 $\times$ $10^{-13}$M.
I can not understand why we have to multiply the corresponding portion of $Cl^-$ by two (In the parenthesis). Is the book making an error or am I missing something ? According to the ending pages of the book the answer is 2.8 $\times$ $10^{-13}$ M and not 2.6302 $\times$ $10^{-13}$M. Please be aware that the differences do not result from approximations.
