4. Calculate the pH of the following solutions a. 0.25 M HF (pKa = 3.14) b. 0.65
ID: 1045194 • Letter: 4
Question
4. Calculate the pH of the following solutions a. 0.25 M HF (pKa = 3.14) b. 0.65 M CH3NH2 (Kb = 4.38×10-4) c. 0. 50 M NH4Cl (Kb of NH3 = 1.8×10-5) d. 0.35 M NaF (Ka of HF = 7.2×10-4)
4. Calculate the pH of the following solutions a. 0.25 M HF (pKa = 3.14) b. 0.65 M CH3NH2 (Kb = 4.38×10-4) c. 0. 50 M NH4Cl (Kb of NH3 = 1.8×10-5) d. 0.35 M NaF (Ka of HF = 7.2×10-4)
4. Calculate the pH of the following solutions a. 0.25 M HF (pKa = 3.14) b. 0.65 M CH3NH2 (Kb = 4.38×10-4) c. 0. 50 M NH4Cl (Kb of NH3 = 1.8×10-5) d. 0.35 M NaF (Ka of HF = 7.2×10-4)
Explanation / Answer
a) HF= 0.25M
Pka= 3.14
-log(Pka) = 3.14
Ka= 10^-3.14 =7.24x10^-4
for weak acids
[H+] = square root of KaxC
[H+]= square root of (7.24x10^-4X0.25)
[H+]= 1.345x10^-2M
-log[H+]= -log(1.345x10^-2)
PH= 1.87
b)
CH3NH2 = 0.65M
Kb= 4.38x10^-4
for weak bases
[OH-] = square root of KbxC
[OH-] = squarer oot of (4.38x10^-4x0.65)
[OH-] = 1.687x10^-2M
-log[OH-]= -log(1.687x10^-2)
POH= 1.77
PH+POH= 14
PH= 14- POH
PH= 14-1.77
PH= 12.23
c)
NH4Cl = 0.50M
NH4Cl ---------------- NH4+ + Cl-
NH4+ + H2O ---------------- NH4OH + H+
0.50 0 0
-x +x +x
0.50-x +x +x
Kb =1.8x10-5
KaxKb= Kw whereKw= 1.0x10^-14
Ka= Kw/Kb = 1.0x10^-14/1.8x10^-5 = 5.56x10^-10
Ka= 5.56x10^-10
Ka=[NH4OH][H+]/[NH4+]
5.56x10^-10 = x*x/(0.50-x)
for solving the equation
x= 1.667x10^-5
[H+]= 1.667x10^-5M
-log[H+]= -log(1.667x10^-5)
PH= 4.78
d)
NaF= 0.35M
NaF------------------ Na+ + F-
F- + H2O----------------- HF + OH-
0.35 0 0
-x +x +x
0.35-x +x +x
Ka= 7.2x10^-4
kb = Kw/Ka= 1.0x10^-14/7.2x10^-4 =1.39x10^-11
Kb= [HF][OH-]/[F-]
1.39x10^=11 = x*x/(0.35-x)
for solving the equation
x= 2.21x10^-6
[OH-] = 2.21x10^-6
-log[OH-]= -log(2.21x10^-6)
POH= 5.66
PH= 14-POH
PH= 14-5.66
PH= 8.34
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