Ionic EquilibriumHard
Question
When 100 ml of 0.4 M CH3COOH are mixed with 100 ml of 0.2 M NaOH, the [H3O+] in the solution is
approximately : [Ka(CH3COOH) = 1.8 × 10-5]
approximately : [Ka(CH3COOH) = 1.8 × 10-5]
Options
A.1.8 × 10-6 M
B.1.8 ×10-5 M
C.9 × 10-6 M
D.9 × 10-5 M.
Solution
CH3COOH + NaOH → CH3COONa + H2O
time t = 0 40 mmole 20 mmole
time t = t 20 mmole - 20 mmole
pH = pKa+ log
⇒ pH = pKa ⇒ [H+] = Ka = 1.8 × 10-5 M
time t = 0 40 mmole 20 mmole
time t = t 20 mmole - 20 mmole
pH = pKa+ log
⇒ pH = pKa ⇒ [H+] = Ka = 1.8 × 10-5 MCreate a free account to view solution
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