Ionic EquilibriumHard
Question
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 added are
Options
A.3.50, 7
B.4.2, 7
C.4.2, 8.1
D.4.2, 8.25
Solution
C6H5COOH + OH- → C6H5COO- + H2O
t = 0 2 1
teq 1 - 1
pH = pKa = 4.2
C6H5COOH + OH- C6H5COO- + H2O
t=0 2 2
teq - - 2
[C6H5COO-] =
= 0.01 M
∴ pH = 7 + pKa +
log C = 7 + 
log (0.01) = 8.1
t = 0 2 1
teq 1 - 1
pH = pKa = 4.2
C6H5COOH + OH- C6H5COO- + H2O
t=0 2 2
teq - - 2
[C6H5COO-] =
= 0.01 M∴ pH = 7 + pKa +
log C = 7 + log (0.01) = 8.1 Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Of the given anions, the strongest Bronsted base is :...2-Hexyne gives trans -2-Hexene on treatment with :-...Two buffer solutions, A and B, each made with acetic acid and sodium acetate differ in their pH by one unit, A has salt ...The pH of 0.1 M solution of the following salts increases in the order:...The solubility product of a salt having general formula MX2, in water is: 4 × 10-12. The concentration of M2+ ions ...