Ionic EquilibriumHard
Question
The pH of a solution obtained by mixing 100 ml of 0.2 M CH3COOH with 100 ml of 0.2 M NaOH would be :
(pKa for CH3COOH = 4.74)
(pKa for CH3COOH = 4.74)
Options
A.4.74
B.8.87
C.9.10
D.8.57
Solution
CH3COOH + OH- CH3COO- + H2O
t = 0 20 20
t = eq - - 20
So, [CH3COO-] =
= 0.1 M
pH = 7 +
pKa + 
log C = 7 + 2.37 + 
log 10-1 = 7 + 2.37 - 0.5 = 8.87
t = 0 20 20
t = eq - - 20
So, [CH3COO-] =
= 0.1 MpH = 7 +
pKa + log C = 7 + 2.37 + log 10-1 = 7 + 2.37 - 0.5 = 8.87Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
A weak acid (HA) after treatment with 12 mL of 0.1 M strong base (BOH) has a pH of 5. At the end point, the volume of sa...Ammonia forms the complex ion [Cu(NH3)4]2+ with copper ions in alkaline solutions but not in acidic solutions. What is t...50 mL of 2N acetic acid mixed with 10 mL of 1N sodium acetate solution will have an approximate pH of (Ka = 10-5)...The pH of a 0.1 molar solution of the acid HQ is 3. The value of the ionization constant, Ka of this acid is :-...Which is/are correct statements : (i) In any strong acid’s solution, the concentration of [OH-] will be zero. (ii) If DG...