What is the pH of a 200ml solution of 0.25M HCl combined with a 100ml solution of 0.25 NaOH?
Given:
M(HCl) = 0.25 M
V(HCl) = 200 mL
M(NaOH) = 0.25 M
V(NaOH) = 100 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.25 M * 200 mL = 50 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.25 M * 100 mL = 25 mmol
We have:
mol(HCl) = 50 mmol
mol(NaOH) = 25 mmol
25 mmol of both will react
remaining mol of HCl = 25 mmol
Total volume = 300.0 mL
[H+]= mol of acid remaining / volume
[H+] = 25 mmol/300.0 mL
= 8.333*10^-2 M
use:
pH = -log [H+]
= -log (8.333*10^-2)
= 1.0792
Answer: 1.08
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