What is the pH of a solution obtained by mixing 50.0 mL of 0.20 M HCl with 50.0 mL of 0.10 M NaOH? Assume volumes are additive and complete dissociation.
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Correct answer
B. 1.30
Principle or equation
pH = -log[H+]. In a strong acid-strong base reaction, the excess H+ concentration is calculated after neutralization.
Why this answer is correct
Moles H+ = 0.050 L * 0.20 M = 0.010 mol. Moles OH- = 0.050 L * 0.10 M = 0.0050 mol. Excess H+ = 0.010 - 0.0050 = 0.0050 mol. Total volume = 0.100 L. [H+] = 0.0050 / 0.100 = 0.050 M. pH = -log(0.050) ≈ 1.30.
Example
If 25.0 mL of 0.10 M HCl is mixed with 25.0 mL of 0.05 M NaOH, excess H+ = 0.0025 - 0.00125 = 0.00125 mol; [H+] = 0.00125/0.050 = 0.025 M; pH = 1.60.
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