Ionic EquilibriumHard
Question
A 40.0 ml solution of weak base, BOH is titrated with 0.1 N – HCl solution. The pH of the solution is found to be 10.0 and 9.0 after adding 5.0 ml and 20.0 ml of the acid, respectively. The dissociation constant of the base is (log 2 = 0.3)
Options
A.2 × 10−5
B.1 × 10−5
C.4 × 10−5
D.5 × 10−5
Solution
$BOH + H^{+} \rightleftharpoons B^{+} + H_{2}O $$$\frac{40 \times C}{40 + V}M\frac{V \times 0.1}{40 + V}M0$$
Final $\frac{40C - 0.1V}{40 + V}0\frac{0.1V}{40 + V}$
H+ must be a limiting reagent because both PH are above >.0.
Now, $14 - 10 = P^{OH} = P^{K_{b}} + \log\frac{0.1V}{40C - 0.1V}$
$14 - 10 = P^{K_{b}} + \log\frac{0.1 \times 5}{40C - 0.1 \times 5}(1) $$$14 - 9 = P^{K_{b}} + \log\frac{0.1 \times 20}{40C - 0.1 \times 20}(2) $$$\therefore K_{b} = 2 \times 10^{- 5}$
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
A solution is 0.1 M CH3COOH and 0.1 M CH3COONa. Which of the following solution will change its pH significantly ?...The pH of the solution obtained on neturalisation of 40 ml 0.1 M NaOH with 40 ml0.1 MCH3 COOH is...To 100 ml of a solution, which contains 8.32 × 10–3 g lead ions, 10−4 moles of H2SO4 is added. How much lead remains in ...Ka for formic acid and acetic acid are 2.1 × 10-4 and 1.1 × 10-5 respectively. The relative strenth of acids i...The species present in solution when CO2 is dissolved in water are...