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During the titration of 50.00 mL of 0.202 M weak acid solution, Ka = 3.76 x...

During the titration of 50.00 mL of 0.202 M weak acid solution, Ka = 3.76 x 10-7, what is the pH after 25.00 mL of 3.76 M NaOH solution has been added?

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Answer #1

HA (aq.) + NaOH ------------> NaA (aq.) + H2O (l)

Initial moles of weak acid = 0.202 * 50.00 / 1000 = 0.0101 mol

Moles of NaOH = 3.76 * 25.00 / 1000 = 0.094 mol

Moles of NaOH > moles of weak acid

So, remaining moles of NaOH = 0.094 - 0.0101 = 0.0839 mol

Final volume = 50.00 + 25.00 = 75.00 mL = 0.07500 L

Final concentration of NaOH = 0.0839 / 0.0750 = 1.12 M

So, pOH = - Log[OH-] = - Log(1.12) = - 0.0492

pH = 14 - pOH = 14 - ( - 0.0492) = 14.0492

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