Question

What is the pH of a 0.18 M CH3NH3+Cl– aqueous solution? The pKb ofCH3NH2 is 3.44

Answer 1

5.65

Step-by-step explanation:

Given that:

`pK_(b)\ of\ CH_3NH_2=3.44`

`K_(b)\ of\ CH_3NH_2=10^(-3.44)=3.6308* 10^(-4)`

`K_a\ of\ CH_3NH_3^+Cl^-=\frac {K_w}{K_b}=\frac {10^(-14)}{3.6308* 10^(-4)}=2.7542* 10^(-11)`

Concentration = 0.18 M

Consider the ICE take for the dissociation as:

`CH_3NH_3^+` ⇄ H⁺ +
`CH_3NH_2`

At t=0 0.18 - -

At t =equilibrium (0.18-x) x x

The expression for dissociation constant of acetic acid is:

`K_(a)=\frac {\left [ H^(+) \right ]\left [CH_3NH_2 \right ]}{[CH_3NH_3^+]}`

`2.7542* 10^(-11)=\frac {x^2}{0.18-x}`

x is very small, so (0.18 - x) ≅ 0.18

Solving for x, we get:

x = 0.2227×10⁻⁵ M

pH = -log[H⁺] = -log(0.2227×10⁻⁵) = 5.65