Ionic EquilibriumMedium
Question
The concentration of acetate ions in 1 M acetic acid (Ka = 2 × 10−5) solution containing 0.1 M HCl is
Options
A.2 × 10–1 M
B.2 × 10–3 M
C.2 × 10–4 M
D.4.4 × 10–3 M
Solution
CH3COOH $\rightleftharpoons$ CH3COO– + H+
1M 0 0.1 M
Equilibrium 1 – x x 0.1 + x
$\simeq$1M $\simeq$0.1 M
Now, $K_{a} = \frac{\left\lbrack CH_{3}COO^{-} \right\rbrack\left\lbrack H^{+} \right\rbrack}{\left\lbrack CH_{3}COOH \right\rbrack} \Rightarrow 2 \times 10^{- 5} = \frac{x \times 0.1}{1}$
⇒ x = 2 × 10–4 M
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Which of the following is an example of addition copolymer :-...Which of the following will show common ion effect and form a buffer solution ?...Which of the following result(s) is/ are correct for the equilibrium state in a solution originally having 0.1 M-CH3COOH...When 0.1 mole solid NaOH is added in 1lt of 0.1M NH3(aq) then which statement is going to wrong?(Kb = 2 × 10-5, log...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 ...