A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 19.0 mL of KOH.
Express your answer numerically.
Given:
M(HBr) = 0.15 M
V(HBr) = 50 mL
M(KOH) = 0.25 M
V(KOH) = 19 mL
mol(HBr) = M(HBr) * V(HBr)
mol(HBr) = 0.15 M * 50 mL = 7.5 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.25 M * 19 mL = 4.75 mmol
We have:
mol(HBr) = 7.5 mmol
mol(KOH) = 4.75 mmol
4.75 mmol of both will react
remaining mol of HBr = 2.75 mmol
Total volume = 69.0 mL
[H+]= mol of acid remaining / volume
[H+] = 2.75 mmol/69.0 mL
= 3.986*10^-2 M
use:
pH = -log [H+]
= -log (3.986*10^-2)
= 1.3995
Answer: 1.40
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