Question

Question 8. Suppose ZnS(s) is saturated with 0.04 M K2S (K2S completely dissolves and ZnS has limited solubility) (Ksp for ZnS is 2x10-25) (a) (2 pts) What is the ionic strength of the solution? Show calculations (b) (4 pts) What are the activity coefficients of each specie?(Please apply Debye Hückel equation given in lecture notes considering the size of the ion if necessary) (c) (4 pts) What is pZn by taking activity coefficients into consideration? Please show calculations

Answer 1

Data:
ZnS_(s) is Saturated with 0,04 M K₂S
Kps (ZnS) = 2.0x10^-25 M

(a) in Solution
ZnS_(s)----------------> Zn²⁺ + S^2- Kps= [Zn²⁺] [S^2-]

K₂S --------------------> 2K⁺ + S^2- 0,04 M in solution

ionic strength of the solution:
$\mu = \frac{1}{2} \sum (C_{a})(Z_{a}^{2})$ $\mu = \frac{1}{2} \sum (C_{a})(Z_{a}^{2}) \;\;\;\;\;\;\;(1)$

$\mu = \frac{1}{2} \sum 2(0,04M) (+1)^{2} + (0,04M)(-2)^{2} \;\;\;\;\;\;\;$

$\mu = 0,12 \;\;\;\;\;\;\;$

(B) What are the activity coefficients of each specie?(Please apply Debye Hückel equation given in lecture notes considering the size of the ion if necessary).

$-log\; \gamma = \frac{0,512z^{2} \sqrt{\mu}}{1+0,329 \alpha \sqrt{\mu}} (2)$

$-log\; \gamma Zn^{2+} = \frac{0,512(2)^{2} \sqrt{\mu}}{1+0,329 \alpha \sqrt{\mu}} \;\;\;\;(2)$

$\alpha Zn^{2+}= 0,60\;nm$

is replaced in 2

$\gamma = 0,197 \;\;\;Zn^{2+}$

$\gamma= 0,297\; S^{2-}$

$\gamma = 0,73\;\;K^{+}$​​​​​​​