Question:
A solution is prepared by adding 0.5 L of 0.5 M NaOH to 0.5 L of 0.55 M formic acid. What is the pH of the resultant solution? (\(K_a\) of formic acid = \(1.8 \times 10^{-4}\); \(\log(1.8) = 0.26\))
A solution is prepared by adding 0.5 L of 0.5 M NaOH to 0.5 L of 0.55 M formic acid. What is the pH of the resultant solution? (\(K_a\) of formic acid = \(1.8 \times 10^{-4}\); \(\log(1.8) = 0.26\))
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Use the Henderson-Hasselbalch equation for buffer solutions created by partially neutralizing a weak acid.
Updated On: Jun 9, 2026
- 3.74
- 4.74
- 2.74
- 3.26
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The Correct Option is B
Solution and Explanation
Concept:
The reaction between a strong base (NaOH) and a weak acid (formic acid, HCOOH) forms a buffer solution of a weak acid and its conjugate base salt (HCOONa) when the base is the limiting reagent.
Step 1: Calculate the moles of reactants.
Moles of NaOH = 0.5 L \(\times\) 0.5 M \[ \text{Moles of NaOH} = 0.25 \text{ mol} \] Moles of Formic Acid = 0.5 L \(\times\) 0.55 M \[ \text{Moles of HCOOH} = 0.275 \text{ mol} \]
Step 2: Determine the composition of the buffer.
The neutralization reaction is: \[ HCOOH + NaOH \rightarrow HCOONa + H_2O \] NaOH is the limiting reagent (0.25 mol). It will react completely with 0.25 mol of HCOOH. \[ \text{Remaining HCOOH} = 0.275 - 0.25 = 0.025 \text{ mol} \] \[ \text{Formed HCOONa (Salt)} = 0.25 \text{ mol} \]
Step 3: Apply the Henderson-Hasselbalch equation.
The formula for the pH of an acidic buffer is: \[ pH = pK_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right) \] Calculate \(pK_a\): \[ pK_a = -\log(1.8 \times 10^{-4}) \] \[ pK_a = 4 - \log(1.8) = 4 - 0.26 = 3.74 \] Calculate the pH: \[ pH = 3.74 + \log\left(\frac{0.25}{0.025}\right) \] \[ pH = 3.74 + \log(10) \] \[ pH = 3.74 + 1 = 4.74 \] \[ \boxed{4.74} \]
Step 1: Calculate the moles of reactants.
Moles of NaOH = 0.5 L \(\times\) 0.5 M \[ \text{Moles of NaOH} = 0.25 \text{ mol} \] Moles of Formic Acid = 0.5 L \(\times\) 0.55 M \[ \text{Moles of HCOOH} = 0.275 \text{ mol} \]
Step 2: Determine the composition of the buffer.
The neutralization reaction is: \[ HCOOH + NaOH \rightarrow HCOONa + H_2O \] NaOH is the limiting reagent (0.25 mol). It will react completely with 0.25 mol of HCOOH. \[ \text{Remaining HCOOH} = 0.275 - 0.25 = 0.025 \text{ mol} \] \[ \text{Formed HCOONa (Salt)} = 0.25 \text{ mol} \]
Step 3: Apply the Henderson-Hasselbalch equation.
The formula for the pH of an acidic buffer is: \[ pH = pK_a + \log\left(\frac{[\text{salt}]}{[\text{acid}]}\right) \] Calculate \(pK_a\): \[ pK_a = -\log(1.8 \times 10^{-4}) \] \[ pK_a = 4 - \log(1.8) = 4 - 0.26 = 3.74 \] Calculate the pH: \[ pH = 3.74 + \log\left(\frac{0.25}{0.025}\right) \] \[ pH = 3.74 + \log(10) \] \[ pH = 3.74 + 1 = 4.74 \] \[ \boxed{4.74} \]
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