When a 17.5 mL sample of a 0.359 M aqueous hypochlorous acid solution is titrated with a 0.460 M aqueous sodium hydroxide solution, what is the pH after 20.5 mL of sodium hydroxide have been added? pH =
Ka of HClO = 3.0*10^-8
Given:
M(HClO) = 0.359 M
V(HClO) = 17.5 mL
M(NaOH) = 0.46 M
V(NaOH) = 20.5 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.359 M * 17.5 mL = 6.2825 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.46 M * 20.5 mL = 9.43 mmol
We have:
mol(HClO) = 6.2825 mmol
mol(NaOH) = 9.43 mmol
6.2825 mmol of both will react
excess NaOH remaining = 3.1475 mmol
Volume of Solution = 17.5 + 20.5 = 38 mL
[OH-] = 3.1475 mmol/38 mL = 0.0828 M
use:
pOH = -log [OH-]
= -log (8.283*10^-2)
= 1.08
use:
PH = 14 - pOH
= 14 - 1.08
= 12.92
Answer: 12.92
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