a)when 11.0 mL of NaOH is added
Given:
M(C3H7COOH) = 0.1 M
V(C3H7COOH) = 20 mL
M(NaOH) = 0.1 M
V(NaOH) = 11 mL
mol(C3H7COOH) = M(C3H7COOH) * V(C3H7COOH)
mol(C3H7COOH) = 0.1 M * 20 mL = 2 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 11 mL = 1.1 mmol
We have:
mol(C3H7COOH) = 2 mmol
mol(NaOH) = 1.1 mmol
1.1 mmol of both will react
excess C3H7COOH remaining = 0.9 mmol
Volume of Solution = 20 + 11 = 31 mL
[C3H7COOH] = 0.9 mmol/31 mL = 0.029M
[C3H7COO-] = 1.1/31 = 0.0355M
They form acidic buffer
acid is C3H7COOH
conjugate base is C3H7COO-
Ka = 1.54*10^-5
pKa = - log (Ka)
= - log(1.54*10^-5)
= 4.812
use:
pH = pKa + log {[conjugate base]/[acid]}
= 4.812+ log {3.548*10^-2/2.903*10^-2}
= 4.9
Answer: 4.90
b)
when 20.4 mL of NaOH is added
Given:
M(C3H7COOH) = 0.1 M
V(C3H7COOH) = 20 mL
M(NaOH) = 0.1 M
V(NaOH) = 20.4 mL
mol(C3H7COOH) = M(C3H7COOH) * V(C3H7COOH)
mol(C3H7COOH) = 0.1 M * 20 mL = 2 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 20.4 mL = 2.04 mmol
We have:
mol(C3H7COOH) = 2 mmol
mol(NaOH) = 2.04 mmol
2 mmol of both will react
excess NaOH remaining = 0.04 mmol
Volume of Solution = 20 + 20.4 = 40.4 mL
[OH-] = 0.04 mmol/40.4 mL = 0.001 M
use:
pOH = -log [OH-]
= -log (9.901*10^-4)
= 3.0043
use:
PH = 14 - pOH
= 14 - 3.0043
= 10.9957
Answer: 11.00
c)
when 29.0 mL of NaOH is added
Given:
M(C3H7COOH) = 0.1 M
V(C3H7COOH) = 20 mL
M(NaOH) = 0.1 M
V(NaOH) = 29 mL
mol(C3H7COOH) = M(C3H7COOH) * V(C3H7COOH)
mol(C3H7COOH) = 0.1 M * 20 mL = 2 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.1 M * 29 mL = 2.9 mmol
We have:
mol(C3H7COOH) = 2 mmol
mol(NaOH) = 2.9 mmol
2 mmol of both will react
excess NaOH remaining = 0.9 mmol
Volume of Solution = 20 + 29 = 49 mL
[OH-] = 0.9 mmol/49 mL = 0.0184 M
use:
pOH = -log [OH-]
= -log (1.837*10^-2)
= 1.736
use:
PH = 14 - pOH
= 14 - 1.736
= 12.264
Answer: 12.26
Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH CH...
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Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K, = 1.54 x 10-5), with 0.1000 M NaOH solution after the following additions of titrant. (a) 10.00 mL: pH = (b) 20.90 mL: pH = (c) 27.00 mL: pH =
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PLEASE SHOW ALL WORK Be sure to answer all parts. Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K. -1.54 x 10-5), with 0.1000 M NaOH solution after the following additions of titrant. (a) 10.00 mL: pH = (b) 20.50 mL: (c) 30.00 mL: