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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.

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    $\begingroup$ Every dissociating mole of ZnCl2 generates 2 mole of Cl- ? $\endgroup$ – Buck Thorn Dec 25 '20 at 15:02
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    $\begingroup$ You're right, the book is wrong. However you have far too many significant figures in your final answer. The Ksp only has one significant figure. // Both you and the book show to few significant figures in the ionic strength calculation. $\endgroup$ – MaxW Dec 25 '20 at 15:53
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    $\begingroup$ The problem is worked correctly in the errata for the book on page 5 of the pdf. cengage.com/resource_uploads/downloads/0495558281_519368.pdf $\endgroup$ – MaxW Dec 25 '20 at 16:43
  • $\begingroup$ @MaxW thank you very much for that document. I wasn't aware of that. It helped me a lot. :) $\endgroup$ – Sam Dec 25 '20 at 17:12

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