Question

The approximate pH of a 0.1 M aqueous $H_{2}S$ solution is ( $K_{1}$ and $K_{2}$ for $H_{2}S$ are $1\times 10^{- 7}$ and $1.3\times 10^{- 13}$ respectively at 25 $^\circ $ C)

• A. 4
• B. 2
• C. 8
• D. 6

Answer 1

Answer: (A)

Second dissociation of $H_{2}S$ is negligible from data

$H_{2}S\rightleftharpoons H^{+}+HS^{-}$

$\left[\right.H^{+}\left]\right.=\left[\right.HS^{-}\left]\right.=x$

$K_{1}=\frac{\left[\right. H^{+} \left]\right. \left[\right. H S^{-} \left]\right.}{\left[\right. H_{2} S \left]\right.}$

$1\times 10^{- 7}=\frac{x^{2}}{0.1}$

So $x=10^{- 4}$

$pH=-log \left[\right.H^{+}\left]\right.$

$=-log 10^{- 4}$

$=4log 10=4$ .