a) 0.01M CH3COONa Weak base CH3COO- + H2O <--------> CH3COOH + OH- Kb = [CH3COOH
ID: 699435 • Letter: A
Question
a) 0.01M CH3COONa
Weak base
CH3COO- + H2O <--------> CH3COOH + OH-
Kb = [CH3COOH][OH-]/[CH3COO-]= 5.71×10^-10
at equillibrium
[CH3COOH] = x
[ OH-] = x
[ CH3COO- ] = 0.01 - x
x^2/0.01-x = 5.71×10^-10
x is small quantity, so we can assume 0.01 - x = 0.01
x^2/0.01 = 5.71×10^-10
x = 2.39×10^-6
[ OH- ] = 2.39 ×10^-6M
pOH = 5.62
pH = 14 - pOH
= 14 - 5.62
= 8.38
b) 0.01M NH3 + 0.01M NH4Cl
Buffer
Henderson - Hasselbalch equation is
pH = pKa + log([A-]/[HA])
= 4.76 + log(0.01M/0.01M)
= 4.76 + 0
= 4.76
c) 0.01M HNO3
strong acid
HNO3 + H2O -------> H3O+ + NO3-
Completely dissociate , so
[HNO3 ] = [H3O+]
pH = -log[H3O+]
= - log ( 0.01)
= 2
d) 0.01M NaOH
strong base
NaOH -------> Na+ + OH-
completely dissociate
[NaOH] = [OH-]
pOH = -log[OH-]
= - log(0.01)
= 2
pH = 14 -2
= 12
e) 0.01M C2H2O4
acid
first dissociation
HOOCCOOH <-------> HOOCCOO- + H+
Ka1 = [HOOCCOO-] [H+] /[HOOCCOOH] = 5.6 × 10^-2
x^2/(0.01- x ) = 5.6×10^-2
x = 0.00866M
[H+] = 0.00866M
[ HOOCCOOO-] = 0.00866M
second dissociation
HOOCCOO- ------------> -OOCCOO- + H+
Ka2 = [-OOCCOO-][H+]/[HOOCCOO-] = 5.42 ×10^-5
x^2/0.00866 - x = 5.42 ×10^-5
x^2/0.00866 = 5.42 ×10^-5
x =
[H+] = 6.85×10^-4M
total [H+] = 0.00866M + 0.000685M = 0.009345M
pH = -log(0.009345)
= 2.03
Explanation / Answer
6. Classify each of the solutions below as to whether it is an acid, a base, a strong acid, a strong base, a weak base, or buffer. In each case, calculate the corresponding pH. (15 points) (a) 0.01 M CH COONa, (b) 0.01 M NH3 + 0.01M NH4CI, (c) 0.01 M HNO3, (d) 0.01 M NaOH, (e) 0.01 M C2H204. (HOOCCOONa)
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