14.5 mL of 0.30M NaOH is titrated with 20.2 mL of 0.25M HCl. What is the...

Question

Question

14.5 mL of 0.30M NaOH is titrated with 20.2 mL of 0.25M HCl.

What is the final pH of the solution?

Answer 1

Answer #1

NaOH + HCl NaCl + H₂O

Millimole of NaOH = molarity × volume (ml) = 0.30 × 14.5 = 4.35

Millimole of HCl = 0.25 × 20.2 = 5.05

Millimole of HCl left = 5.05 - 4.35 = 0.7

Total volume of solution = 14.5ml + 20.2ml = 34.7ml

Molarity of HCl = 0.7 mmol/34.7ml = 0.0201729M

pH = - log[H⁺] = - log (0.0201729) = 1.695 (answer)

Answer 2

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