Ionic EquilibriumHard
Question
An amount of 0.1 mole of CH3NH2 (Kb = 5 × 10–4) is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H+ concentration in the solution?
Options
A.1.25 × 10–4 M
B.8 × 10–11 M
C.1.6 × 10–11 M
D.2 × 10–3 M
Solution
CH3NH2 + H+ $\rightleftharpoons$ CH3NH3+
0.1 mole 0.08 mole 0
Final 0.02 mole 0 0.08 mole
$P^{OH} = P^{K_{b} + \log\frac{\left\lbrack CH_{3}NH_{3}^{+} \right\rbrack_{0}}{\left\lbrack CH_{3}NH_{2} \right\rbrack_{0}}} = - \log\left( 5 \times 10^{- 4} \right) + \log\frac{0.08}{0.02} $$$\therefore\left\lbrack OH^{-} \right\rbrack = \frac{5 \times 10^{- 4}}{4} \Rightarrow \left\lbrack H^{+} \right\rbrack = \frac{10^{- 14}}{\left\lbrack OH^{-} \right\rbrack} = 8 \times 10^{- 11}\text{ M}$$
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Which of the following is an example of addition copolymer :-...The amino acid glycine (NH2CH2COOH) is basic because of its –NH2 group and acidic because of its –COOH group. By a proce...50 mL of 0.1 M NaOH is added to 75 mL of 0.1M NH4Cl to make a basic buffer. If pKa of NH4+ is 9.26 then pH will be :-...Which of the following statements about a weak acid strong base titration is/are correct ?...Find the number of P - O - P linkage in (HPO3)n where n = 3...