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

Answer 2

The pH of a solution will be "11.568".

Methylammonium chloride:

By using Henderson Hassalbalch's equation, we get

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

pH = 14-pOH

Now,

pOH will be:

→ $pOH = 3.43+(-0.397)$

           $= 3.03$

→ $pH = 14-pOH$

         $= 14-3.03$

         $= 10.97$

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

After adding base, we get

→ $[Salt] = \frac{Molarity\times Volume}{Final \ Volume}$

            $= \frac{0.1\times 20}{20+50}$

            $= 0.0286 \ M$

[base] = [methylamine] = 0.10

After mixing with salt, we get

→ $[base] = \frac{Molarity\times Volume }{Final \ volume}$

            $= \frac{0.1\times 50}{20+50}$

            $= 0.0714 \ M$

→ pKb = -log[Kb] = 3.43

By substituting the values,

→ $pOH = 3.43+log \frac{0.0286}{0.0714}$

            $= 2.432$

hence,

→  pH = 14-pOH

         $= 14-2.432$

         $= 11.568$

Thus the above answer is correct.

Find out more information about methylammonium chloride here:

https://brainly.com/question/522383