Question

What is the minimum $pH$ required to prevent the precipitation of $ZnS$ in a solution that is $0.01 \, M \, Z n C l_{2}$ and saturated with $0.10 \, M \, H_{2}S$ ?
[Given $K_{s p}=10^{- 21}, \, K_{a_{1}} \, \times \, K_{a_{2}}=10^{- 20}$ ]

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (B)

$K _{ sp }$ for $ZnS =\left[ Zn ^{2+}\right]\left[ S ^{2-}\right]$
$\left[ S ^{2-}\right]=\frac{10^{-21}}{0.01}=10^{-19}$
$H _{2} S \rightleftharpoons H ^{+}+ HS ^{-} K _{ a _{1}}=\frac{ H ^{+} HS ^{-}}{ H _{2} S } HS ^{-}$
$\rightleftharpoons H ^{+}+ S ^{2-} K _{ a _{2}}=\frac{ H ^{+} S ^{2-}}{ HS ^{-}}$
for $K _{ a _{1}} \cdot K _{ a _{2}}=\frac{\left[ H ^{+}\right]^{2}\left[ s ^{2-}\right]}{\left[ H _{2} s \right]}$
$10^{-20}=\frac{\left[ H ^{+}\right]^{2} \times 10^{-19}}{0.1} \Rightarrow\left[ H ^{+}\right]=0.1$
or $pH =1$