The pH of the diluted solution is 5.53.
Henderson–Hasselbalch equation for buffer solution:
pH = pKa + log(c(CH₃COO⁻)/c(CH₃COOH))
pH = 5.0 ⇒ c(H⁺) = 10⁻⁵ M
pKa = 4.75 ⇒ Ka = 1.78×10⁻⁵
log(cs/ck) = 5.0 - 4.75 = 0.25
c(CH₃COO⁻)/c(CH₃COO⁻) = 10∧(0.25) = 1.778
c(CH₃COO⁻) + c(CH₃COOH) = 0.1 mM
c(CH₃COO⁻) + c(CH₃COOH) = 1 × 10⁻⁴M
c(CH₃COOH) = 3.6x10⁻⁵ M; concentration of acetic acid before dilution
c(CH₃COOH) = 3.6x10⁻⁶ M; concentration of acetic acid after 10 times dilution
c(CH₃COO⁻) = 6.4x10⁻⁵ M; concentration of sodium acetate before dilution
c(CH₃COO⁻) = 6.4x10⁻⁶ M; concentration of sodium acetate after dilution
c(H⁺) = 10⁻⁶ M; concentration of hydrogen ions
After buffer solution is diluted to 1 l:
Ka = (c(CH₃COO⁻) + x) × (c(H⁺) + x) / (c(CH₃COOH) - x)
1.78×10⁻⁵ = (6.4x10⁻⁶ M + x) × (10⁻⁶ M + x) / (3.6x10⁻⁶ M - x)
x = 2.12×10⁻⁶ M
c(H⁺) = 10⁻⁶ M + 2.12×10⁻⁶ M
c(H⁺) = 3.12×10⁻⁶ M
pH = -logc(H⁺)
pH = -log3.12×10⁻⁶ M
pH = 5.53
More about diluted solution: brainly.com/question/24881505
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