Questions: Use the References to access important values if needed for this question. When a 27.4 mL sample of a 0.329 M aqueous hydrocyanic acid solution is titrated with a 0.452 M aqueous potassium hydroxide solution, what is the pH after 29.9 mL of potassium hydroxide have been added? pH=

Solution Steps

First, calculate the initial moles of hydrocyanic acid (HCN) in the solution. Use the formula:

\[ \text{moles of HCN} = \text{volume (L)} \times \text{molarity (M)} \]

Given:

• Volume of HCN = 27.4 mL = 0.0274 L
• Molarity of HCN = 0.329 M

\[ \text{moles of HCN} = 0.0274 \, \text{L} \times 0.329 \, \text{M} = 0.0090146 \, \text{mol} \]

Next, calculate the moles of potassium hydroxide (KOH) added to the solution. Use the formula:

\[ \text{moles of KOH} = \text{volume (L)} \times \text{molarity (M)} \]

Given:

• Volume of KOH = 29.9 mL = 0.0299 L
• Molarity of KOH = 0.452 M

\[ \text{moles of KOH} = 0.0299 \, \text{L} \times 0.452 \, \text{M} = 0.0135148 \, \text{mol} \]

The reaction between HCN and KOH is:

\[ \text{HCN} + \text{KOH} \rightarrow \text{KCN} + \text{H}_2\text{O} \]

Since KOH is in excess, all HCN will react. Calculate the remaining moles of KOH:

\[ \text{Remaining moles of KOH} = \text{moles of KOH} - \text{moles of HCN} = 0.0135148 \, \text{mol} - 0.0090146 \, \text{mol} = 0.0045002 \, \text{mol} \]

The remaining KOH contributes to the concentration of OH\(^-\) ions. Calculate the concentration of OH\(^-\) ions in the total volume of the solution:

Total volume = 27.4 mL + 29.9 mL = 57.3 mL = 0.0573 L

\[ [\text{OH}^-] = \frac{\text{Remaining moles of KOH}}{\text{Total volume (L)}} = \frac{0.0045002 \, \text{mol}}{0.0573 \, \text{L}} = 0.07855 \, \text{M} \]

Calculate the pOH using the concentration of OH\(^-\):

\[ \text{pOH} = -\log_{10}([\text{OH}^-]) = -\log_{10}(0.07855) = 1.104 \]

Finally, calculate the pH:

\[ \text{pH} = 14 - \text{pOH} = 14 - 1.104 = 12.896 \]

Final Answer