Answer
$[H_2S ] = 0.10 \space M$
$[HS^-] = 9 \times 10^{-5} \space M$
$[S^{2-}] = 1 \times 10^{-17} \space M$
$[H_3O^+] = 9 \times 10^{-5} $
$[OH^-] = 1 \times 10^{-10} \space M$
$pH = 4.0 $
$pOH = 10.0$
Work Step by Step
First dissociation:
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ H_2S ]& [ HS^- ]& [ H_3O^+ ]\\
Initial& 0.10 & 0 & 0 \\
Change& -x& +x& +x\\
Equilibrium& 0.10 -x& 0 +x& 0 +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each concentration is equal to its balance coefficient.
$$K_C = \frac{[Products]}{[Reactants]} = \frac{[ HS^- ][ H_3O^+ ]}{[ H_2S ]}$$
$$K_a = \frac{(x)(x)}{[ H_2S ]_{initial} - x}$$
3. Assuming $ 0.10 \gt\gt x:$
$$K_a = \frac{x^2}{[ H_2S ]_{initial}}$$
$$x = \sqrt{K_a \times [ H_2S ]_{initial}} = \sqrt{ 9 \times 10^{-8} \times 0.10 }$$
$x = 9. \times 10^{-5} $
4. Test if the assumption was correct:
$$\frac{ 9. \times 10^{-5} }{ 0.10 } \times 100\% = 0.09 \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 9 \times 10^{-5} $
$[H_2S ] =0.10 - 9 \times 10^{-5} = 0.10 \space M$
---------------
Second dissociation:
1. Draw the ICE table for this equilibrium:
$$\begin{vmatrix}
Compound& [ HS^- ]& [ S^{2-} ]& [ H_3O^+ ]\\
Initial& 9 \times 10^{-5} & 0 & 9 \times 10^{-5} \\
Change& -x& +x& +x\\
Equilibrium& 9 \times 10^{-5} -x& 0 +x& 9 \times 10^{-5} +x\\
\end{vmatrix}$$
2. Write the expression for $K_a$, and substitute the concentrations:
- The exponent of each pressure is equal to its balance coefficient.
$$K_a = \frac{P_{Products}}{P_{Reactants}} = \frac{[ S^{2-} ][ H_3O^+ ]}{[ HS^- ]}$$
$$K_a = \frac{(x)( 9 \times 10^{-5} + x)}{[ HS^- ]_{initial} - x}$$
3. Assuming $ \space 9 \times 10^{-5} \gt\gt x:$
$$K_a = \frac{(x)( 9 \times 10^{-5} )}{[ HS^- ]_{initial}}$$
$$x = \frac{K_a \times [ HS^- ]_{initial}}{ 9 \times 10^{-5} } = \frac{ 1 \times 10^{-17} \times 9 \times 10^{-5} }{ 9 \times 10^{-5} } $$
$x = 1.0 \times 10^{-17} $
4. Test if the assumption was correct:
$$\frac{ 1.0 \times 10^{-17} }{ 9 \times 10^{-5} } \times 100\% = 1.0 \times 10^{-11} \%$$
5. The percent is less than 5%. Thus, it is correct to say that $x = 1.0 \times 10^{-17} $
6. Determine the hydronium concentration: $$[H_3O^+] = x + [H_3O^+]_{initial} = 1.0 \times 10^{-17} + 9 \times 10^{-5} = 9. \times 10^{-5} $$
7. Calculate the pH:
$$pH = -log[H_3O^+] = -log( 9. \times 10^{-5} ) = 4.0 $$
$[HS^-] = 9 \times 10^{-5} - 1 \times 10^{-17} = 9 \times 10^{-5} \space M$
$[S^{2-}] = 1 \times 10^{-17} \space M$
$$[OH^-] = \frac{1.0 \times 10^{-14}}{[H_3O^+]} = \frac{1.0 \times 10^{-14}}{ 9. \times 10^{-5} } = 1 \times 10^{-10} \space M$$
$$pOH = 14.00 - 4.0 = 10.0$$
