Ionic EquilibriumHard
Question
What is [H+] in mol/L of a solution that is 0.20 M in CH3COONa and 0.10 M in CH3COOH ?
(Ka for CH3COOH = 1.8 × 10-5 )
(Ka for CH3COOH = 1.8 × 10-5 )
Options
A.3.5 × 10-4
B.1.1 × 10-5
C.1.8 × 10-5
D.9.0 × 10-6
Solution
CH3COOH (weak acid) and CH3COONa (conjugated salt) from acidic buffer and for acidic buffer,
pH = - log Ka + log
[∵ pKa = - log Ka]
= - log (1.8 × 10-5) + log
= 4.74 + log 2
= 4.74 + 0.3010 = 5.041
Now, [ H+ ] = antilog (- 5.045)
= 9.0 × 10-6 mol/L
pH = - log Ka + log
[∵ pKa = - log Ka]= - log (1.8 × 10-5) + log
= 4.74 + log 2
= 4.74 + 0.3010 = 5.041
Now, [ H+ ] = antilog (- 5.045)
= 9.0 × 10-6 mol/L
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
pH of 0.01 M HS- will be:...A certain indicator (an organic dye) has pKa = 5. For which of the following titrations may it be suitable?...100 mL of 0.02 M benzoic acid (pKa = 4.2) is titrated using 0.02 M NaOH. pH after 50 mL and 100 mL of NaOH have been add...10 mL, of a strong acid solution of pH = 2.000 are mixed with 990 mL of another strong acid solution of pH = 4.000. The ...A volume of 18 ml of acetic acid mixture and sodium acetate required 6 ml of 0.1 M-NaOH for neutralization of the acid a...