Calculate the pH and [H3O+] of the following solutions: a) 0.20 M potassium hydr
ID: 1073224 • Letter: C
Question
Calculate the pH and [H3O+] of the following solutions: a) 0.20 M potassium hydroxide (KOH, strong base) b) 0.10 M propanoic acid (CH3CH2COOH, Ka = 1.3 x 10-5) c) 1.50 M diethyl amine ((CH3)2NH, Kb = 5.9 x 10-4) Calculate the pH and [H3O+] of the following solutions: a) 0.20 M potassium hydroxide (KOH, strong base) b) 0.10 M propanoic acid (CH3CH2COOH, Ka = 1.3 x 10-5) c) 1.50 M diethyl amine ((CH3)2NH, Kb = 5.9 x 10-4) Calculate the pH and [H3O+] of the following solutions: a) 0.20 M potassium hydroxide (KOH, strong base) b) 0.10 M propanoic acid (CH3CH2COOH, Ka = 1.3 x 10-5) c) 1.50 M diethyl amine ((CH3)2NH, Kb = 5.9 x 10-4) Calculate the pH and [H3O+] of the following solutions: a) 0.20 M potassium hydroxide (KOH, strong base) b) 0.10 M propanoic acid (CH3CH2COOH, Ka = 1.3 x 10-5) c) 1.50 M diethyl amine ((CH3)2NH, Kb = 5.9 x 10-4)Explanation / Answer
a. KOH ---------> K+ + OH-
0.2M 0.2M
[OH-] = [KOH]
[OH-] = 0.2M
[H3O+] = Kw/[OH-]
= 1*10-14/0.2
= 5*10-14 M
PH = -log[H3O+]
= -log5*10-14
= 13.3010
b. CH3CH2COOH + H2O ---------> CH3CH2COO- + H3O+
I 0.1 0 0
C -x +x +x
E 0.1-x +x +x
Ka = [CH3CH2COO-][H3O+]/[CH3CH2COOH]
1.3*10-5 = x*x/0.1-x
1.3*10-5 *(0.1-x) = x2
x = 0.001133
[H3O+] = x = 0.001133M
PH = -log[H3O+]
= -log0.001133
= 2.9457
c. (CH3)2NH + H2O ----------> (CH3)2NH2^+ + OH^-
I 1.5 0 0
C -x +x +x
E 1.5-x +x +x
Kb = [ (CH3)2NH2^+] [OH^-]/[(CH3)2NH]
5.9*10-4 = x*x/1.5-x
5.9*10-4 *(1.5-x) = x2
x = 0.02945
[OH^-] = x = 0.02945M
[H3O+] = Kw/[OH^-]
= 1*10^-14/0.02945
= 3.395*10^-13 M
PH = -log[H3O+]
= -log3.395*10^-13
= 12.4691
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