Question

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M methylamine, CH3NH2, with 20.00 mL of 0.10 M methylammonium chloride, CH3NH3Cl? Assume that the volume of the solutions are additive and that Kb = 3.70 × 10-4 for methylamine

Answer 1

Answer:

pH=10.97

Explanation:

the solution of methyl amine with methylammonium chloride will make a buffer solution.

The pH of buffer solution can be obtained using Henderson Hassalbalch's equation, which is:

$pOH=pKb+log\frac{[salt]}{[base]}$

pH = 14- pOH

Let us calculate pOH$pOH=3.43+ (-0.397)=3.03$

$pH=14-pOH=14-3.03=10.97$

[Salt] = [methylammonium chloride] = 0.10 M (initial)

After adding base

$[salt] = \frac{molarityXvolume}{finalvolume}=\frac{0.1X20}{(20+50)}= 0.0286M$

[base] = [Methylamine]=0.10

After mixing with salt

$[base]= \frac{molarityXvolume}{finalvolume}=\frac{0.1X50}{(20+50)}= 0.0714M$

pKb= -log[Kb]= 3.43

Putting values

pOH = $3.43+log(\frac{[0.0286]}{0.0714}$