Question:
Find the pH value of the mixture containing 50 cc M HCl and 30 cc M NaOH solution, assuming both to be completely ionised.
Find the pH value of the mixture containing 50 cc M HCl and 30 cc M NaOH solution, assuming both to be completely ionised.
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For strong acid-strong base titration, the pH after neutralization is calculated from the excess H\(^+\) or OH\(^-\) concentration.
Updated On: May 2, 2026
- 0.7051
- 0.6021
- 10.051
- 8.052
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
This is a strong acid-strong base neutralization. The pH is determined by the concentration of H\(^+\) ions remaining after reaction.
Step 2: Detailed Explanation:
Volume of HCl = 50 cc = 0.05 L, Molarity = 1 M. Moles of H\(^+\) = \(0.05 \times 1 = 0.05\) mol. Volume of NaOH = 30 cc = 0.03 L, Molarity = 1 M. Moles of OH\(^-\) = \(0.03 \times 1 = 0.03\) mol. Reaction: \(H^+ + OH^- \rightarrow H_2O\). Moles of H\(^+\) remaining = \(0.05 - 0.03 = 0.02\) mol. Total volume = \(0.05 + 0.03 = 0.08\) L. \([H^+] = \frac{0.02}{0.08} = 0.25\) M. pH = \(-\log [H^+] = -\log (0.25) = -\log (2.5 \times 10^{-1}) = -(\log 2.5 - 1) = 1 - 0.3979 = 0.6021\).
Step 3: Final Answer:
pH = 0.6021, option (B).
This is a strong acid-strong base neutralization. The pH is determined by the concentration of H\(^+\) ions remaining after reaction.
Step 2: Detailed Explanation:
Volume of HCl = 50 cc = 0.05 L, Molarity = 1 M. Moles of H\(^+\) = \(0.05 \times 1 = 0.05\) mol. Volume of NaOH = 30 cc = 0.03 L, Molarity = 1 M. Moles of OH\(^-\) = \(0.03 \times 1 = 0.03\) mol. Reaction: \(H^+ + OH^- \rightarrow H_2O\). Moles of H\(^+\) remaining = \(0.05 - 0.03 = 0.02\) mol. Total volume = \(0.05 + 0.03 = 0.08\) L. \([H^+] = \frac{0.02}{0.08} = 0.25\) M. pH = \(-\log [H^+] = -\log (0.25) = -\log (2.5 \times 10^{-1}) = -(\log 2.5 - 1) = 1 - 0.3979 = 0.6021\).
Step 3: Final Answer:
pH = 0.6021, option (B).
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