Calculate the pH during the titration of 20.00 mL of 0.1000 M CH3CH2COOH(aq) with 0.2000 M LiOH(aq) after 7 mL of the base have been added. Ka of propanoic acid = 1.3 x 10-5.
Given:
M(CH3CH2COOH) = 0.1 M
V(CH3CH2COOH) = 20 mL
M(LiOH) = 0.2 M
V(LiOH) = 7 mL
mol(CH3CH2COOH) = M(CH3CH2COOH) * V(CH3CH2COOH)
mol(CH3CH2COOH) = 0.1 M * 20 mL = 2 mmol
mol(LiOH) = M(LiOH) * V(LiOH)
mol(LiOH) = 0.2 M * 7 mL = 1.4 mmol
We have:
mol(CH3CH2COOH) = 2 mmol
mol(LiOH) = 1.4 mmol
1.4 mmol of both will react
excess CH3CH2COOH remaining = 0.6 mmol
Volume of Solution = 20 + 7 = 27 mL
[CH3CH2COOH] = 0.6 mmol/27 mL = 0.0222M
[CH3CH2COO-] = 1.4/27 = 0.0519M
They form acidic buffer
acid is CH3CH2COOH
conjugate base is CH3CH2COO-
Ka = 1.3*10^-5
pKa = - log (Ka)
= - log(1.3*10^-5)
= 4.886
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.886+ log {5.185*10^-2/2.222*10^-2}
= 5.254
Answer: 5.25
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