Salt Hydrolysis: Calculation of Solution pH Weak Acid Weak Base CH3COOH (Acetic
ID: 1029054 • Letter: S
Question
Salt Hydrolysis: Calculation of Solution pH Weak Acid Weak Base CH3COOH (Acetic Acid) C6HsCOOH (Benzoic Acid) CH3CH2CH2COOH (Butanoic Acid) HCOOH (Formic Acid) HBrO (Hypobromous Acid) HNO2 (Nitrous Acid) HCIO (Hypochlorous Acid) CH3CH2COOH (Propanoic Acid) HCN (Hydrocyanic Acid) Ka 1.8 x 10 NH3 (Ammonia) 6.5 x 105 C6HNH2 (Aniline) 1.5 x 10-5 (CH3CH2)2NH (Diethyl amine) 1.8 X 104 C5HSN (Pyridine) 2.8 x 10-9 CH3CH2NH2 (Ethyl amine) 4.6 x 10-4 (CH3)3N (Trimethyl amine) 2.9 x 10-8 (CH3)2NH (Dimethyl amine) 1.3 X 105 4.9 X 10-10 1.76 x 10s 3.98 10.10 6.9 x 104 1.7 X 109 5.6 X 104 6.4 x 10 5.4 x 104 Using the table of K values provided above, what is the Kb value for CH3CH2CH2COO7 What is the pKb for this anion? 9.22 What is the pH of a 0.50 M aqueous solution of potassium butanoate? 9.2Explanation / Answer
CH3-CH2-CH2-COOH(aq) -----------> CH3CH2-CH2COO^- (aq) + H^+ (aq)
Ka of CH3-CH2-CH2-COOH = 1.5*10^-5
Kb of CH3-CH2-CH2-COO^- = Kw/Ka
= 1*10^-14/1.5*10^-5 = 6.67*10^-10
PKb = -logKb
= -log6.67*10^-10
= 9.1758
CH3-CH2-CH2-COOK(aq) ----------------> CH3-CH2-CH2-COO^- (aq) + K^+ (aq)
0.5M 0.5M
CH3-CH2-CH2-COO^- (aq) + H2O(l) ------------> CH3-CH2-CH2-COOH (aq) + OH^- (aq)
I 0.5 0 0
C -x +x +x
E 0.5-x +x +x
Kb = [ CH3-CH2-CH2-COOH][OH^-]/[CH3-CH2-CH2-COO^-]
6.67*10^-10 = x*x/0.5-x
6.67*10^-10 *(0.5-x) = x^2
x = 1.83*10^-5
[OH^-] = x = 1.83*10^-5 M
POH = -log[OH^-]
= -log1.83*10^-5
= 4.7375
PH = 14-POH
= 14-4.7375 = 9.2625
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