Question:
If 200 mL of a 0.031 M solution of H$_2$SO$_4$ are added to 84 mL of a 0.150 M KOH solution, what is the pH of the resulting solution?
If 200 mL of a 0.031 M solution of H$_2$SO$_4$ are added to 84 mL of a 0.150 M KOH solution, what is the pH of the resulting solution?
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Always check excess after neutralization.
Updated On: Apr 23, 2026
- 12.4
- 1.7
- 2.2
- 10.9
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The Correct Option is D
Solution and Explanation
Concept:
Acid-base neutralization + excess OH$^-$ determines pH
Step 1: Moles of H$_2$SO$_4$: \[ 0.031 \times 0.2 = 0.0062\ \text{mol} \]
Step 2: H$_2$SO$_4$ gives 2H$^+$: \[ \text{Total H}^+ = 2 \times 0.0062 = 0.0124\ \text{mol} \]
Step 3: Moles of KOH: \[ 0.150 \times 0.084 = 0.0126\ \text{mol} \]
Step 4: Excess OH$^-$: \[ 0.0126 - 0.0124 = 0.0002\ \text{mol} \]
Step 5: [OH$^-$] concentration: \[ \frac{0.0002}{0.284} \approx 7.04 \times 10^{-4} \]
Step 6: pOH and pH: \[ \text{pOH} \approx 3.15 \Rightarrow \text{pH} \approx 10.85 \approx 10.9 \] Conclusion:
pH = 10.9
Step 1: Moles of H$_2$SO$_4$: \[ 0.031 \times 0.2 = 0.0062\ \text{mol} \]
Step 2: H$_2$SO$_4$ gives 2H$^+$: \[ \text{Total H}^+ = 2 \times 0.0062 = 0.0124\ \text{mol} \]
Step 3: Moles of KOH: \[ 0.150 \times 0.084 = 0.0126\ \text{mol} \]
Step 4: Excess OH$^-$: \[ 0.0126 - 0.0124 = 0.0002\ \text{mol} \]
Step 5: [OH$^-$] concentration: \[ \frac{0.0002}{0.284} \approx 7.04 \times 10^{-4} \]
Step 6: pOH and pH: \[ \text{pOH} \approx 3.15 \Rightarrow \text{pH} \approx 10.85 \approx 10.9 \] Conclusion:
pH = 10.9
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