The given problem involves a buffer solution containing hydrogen sulfide ($H_2S$) and its conjugate base ($HS^-$), with a pH change after adding sodium hydrogen sulfide ($NaHS$). The problem asks for the concentration of the added salt.

Given data:

• Initial volume of the buffer solution = $2$ L

• Initial pH = $6.6$

• pH change = $2$ units (final pH = $6.6 + 2 = 8.6$)

• pKa of $H_2S$ = $7$

• Buffer ratio ($\frac{\text{base}}{\text{acid}}$) = $2.5$

• Henderson-Hasselbalch equation:

  [ \text{pH} = \text{pKa} + \log \left( \frac{\text{[Base]}}{\text{[Acid]}} \right) ]

  Substituting values:

  [ 8.6 = 7 + \log \left( \frac{[HS^-]}{[H_2S]} \right) ]

  [ \log \left( \frac{[HS^-]}{[H_2S]} \right) = 1.6 ]

  [ \frac{[HS^-]}{[H_2S]} = 10^{1.6} \approx 40 ]

  Since the original ratio was $2.5$, the added amount of $HS^-$ increased this ratio.

• Given that the added salt concentration is equal to $[HS^-]$:

  $[HS^-] = 2.5 \times 10^{-3} \text{ M}$

Answer:

Concentration of the added NaHS = $2.5 \times 10^{-3}$ mol/L.

Would you like a more detailed explanation? Here are five related questions to expand on this concept:

1. How does the Henderson-Hasselbalch equation help in buffer calculations?
2. Why does adding $NaHS$ increase the pH of the solution?
3. What are the roles of conjugate acid-base pairs in maintaining buffer pH?
4. How can we determine the amount of solid $NaHS$ added in grams?
5. How would the buffer capacity change if a strong acid was added instead?

Tip: When solving buffer problems, always check if the pH change is reasonable based on the given concentrations and buffer capacity.