# Problem in the calculation of ionic strength of a solution

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.

• Every dissociating mole of ZnCl2 generates 2 mole of Cl- ? – Buck Thorn Dec 25 '20 at 15:02
• 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. – MaxW Dec 25 '20 at 15:53
• 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 – MaxW Dec 25 '20 at 16:43
• @MaxW thank you very much for that document. I wasn't aware of that. It helped me a lot. :) – Sam Dec 25 '20 at 17:12