a)when 12.0 mL of HCl is added
Given:
M(HCl) = 0.1 M
V(HCl) = 12 mL
M((C2H5)3N) = 0.1 M
V((C2H5)3N) = 20 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.1 M * 12 mL = 1.2 mmol
mol((C2H5)3N) = M((C2H5)3N) * V((C2H5)3N)
mol((C2H5)3N) = 0.1 M * 20 mL = 2 mmol
We have:
mol(HCl) = 1.2 mmol
mol((C2H5)3N) = 2 mmol
1.2 mmol of both will react
excess (C2H5)3N remaining = 0.8 mmol
Volume of Solution = 12 + 20 = 32 mL
[(C2H5)3N] = 0.8 mmol/32 mL = 0.025 M
[(C2H5)3NH+] = 1.2 mmol/32 mL = 0.0375 M
They form basic buffer
base is (C2H5)3N
conjugate acid is (C2H5)3NH+
Kb = 5.2*10^-4
pKb = - log (Kb)
= - log(5.2*10^-4)
= 3.284
use:
pOH = pKb + log {[conjugate acid]/[base]}
= 3.284+ log {3.75*10^-2/2.5*10^-2}
= 3.46
use:
PH = 14 - pOH
= 14 - 3.4601
= 10.5399
Answer: 10.54
b)when 20.6 mL of HCl is added
Given:
M(HCl) = 0.1 M
V(HCl) = 20.6 mL
M((C2H5)3N) = 0.1 M
V((C2H5)3N) = 20 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.1 M * 20.6 mL = 2.06 mmol
mol((C2H5)3N) = M((C2H5)3N) * V((C2H5)3N)
mol((C2H5)3N) = 0.1 M * 20 mL = 2 mmol
We have:
mol(HCl) = 2.06 mmol
mol((C2H5)3N) = 2 mmol
2 mmol of both will react
excess HCl remaining = 0.06 mmol
Volume of Solution = 20.6 + 20 = 40.6 mL
[H+] = 0.06 mmol/40.6 mL = 0.0015 M
use:
pH = -log [H+]
= -log (1.478*10^-3)
= 2.8304
Answer: 2.83
c)when 29.0 mL of HCl is added
Given:
M(HCl) = 0.1 M
V(HCl) = 29 mL
M((C2H5)3N) = 0.1 M
V((C2H5)3N) = 20 mL
mol(HCl) = M(HCl) * V(HCl)
mol(HCl) = 0.1 M * 29 mL = 2.9 mmol
mol((C2H5)3N) = M((C2H5)3N) * V((C2H5)3N)
mol((C2H5)3N) = 0.1 M * 20 mL = 2 mmol
We have:
mol(HCl) = 2.9 mmol
mol((C2H5)3N) = 2 mmol
2 mmol of both will react
excess HCl remaining = 0.9 mmol
Volume of Solution = 29 + 20 = 49 mL
[H+] = 0.9 mmol/49 mL = 0.0184 M
use:
pH = -log [H+]
= -log (1.837*10^-2)
= 1.736
Answer: 1.74
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