I\'m stuck on a weak acid-strong base titration problem... A titration of .1000M

Question

I'm stuck on a weak acid-strong base titration problem...
A titration of .1000M HOCl is done using .1003M NaOH. If25 mL aliquot of acid is used, determine the volume of NaOHrequired to reach the equivalence point, the half eq. point and pHat the half eq. point. The Ka for HOCl is 3.5 EE -8.
Using ICE I got x^2 / .1 = 3.5 EE -8 so x= 5.92 EE-5 with ainitial pH of 4.23, however I can't seem to make it any further. Since it is the same M for each, does it take the same amount(25mL) of NaOH to reach eq. point or am I missing something?
A titration of .1000M HOCl is done using .1003M NaOH. If25 mL aliquot of acid is used, determine the volume of NaOHrequired to reach the equivalence point, the half eq. point and pHat the half eq. point. The Ka for HOCl is 3.5 EE -8.
Using ICE I got x^2 / .1 = 3.5 EE -8 so x= 5.92 EE-5 with ainitial pH of 4.23, however I can't seem to make it any further. Since it is the same M for each, does it take the same amount(25mL) of NaOH to reach eq. point or am I missing something?

Explanation / Answer

It should take the same amount of NaOH to reach theequivalence point, but the molar concentration are not exactly thesame. To reach the equivalence point: (0.025 L acid)(0.100 mole acid/1 L acid)(1 mole NaOH/1 moleacid)(1 L NaOH/0.1003 mole NaOH) = 0.0249 L NaOH solution The half equivalence point is half of the calculated amountfor the equivalence point: 0.0249 L/2 = 0.0125 L At the half equivalence point, [HA] = [A-]. Use the Henderson-Hasselbach equation to find the pH. pH = pKa + log([A-]/[HA]) pH = pKa + log1 pH = pKa pKa = -log(3.5*10-8) = 7.46 pH = 7.46

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