Question

How many moles of CH₃NH₃Cl need to be added to 200.0 mL of a 0.500 M solution of CH₃NH₂ (Kb for CH₃NH₂ is 4.4 × 10⁻⁴) to make a buffer with a pH of 9.70?

Answer 1

Answer : The moles of weak base $(CH_3NH_2)$ needed to added is, 0.871 mol.

Explanation : Given,

$K_b=4.4\times 10^{-4}$

Concentration of weak base = 5.00 M

Volume of solution = 200.0 mL

pH = 9.70

First we have to calculate the value of $pK_b$.

The expression used for the calculation of $pK_b$ is,

$pK_b=-\log (K_b)$

Now put the value of $K_a$ in this expression, we get:

$pK_b=-\log (4.4\times 10^{-4})$

$pK_b=4-\log (4.4)$

$pK_b=3.36$

Now we have to calculate the value of pOH.

$pOH=14-pH\\\\pOH=14-9.70\\\\pOH=4.3$

Now we have to calculate the moles of weak base $(CH_3NH_2)$.

$\text{Moles of }CH_3NH_2=\text{Concentration of }CH_3NH_2\times \text{Volume of solution}=0.500mole/L\times 0.200L=0.1mole$

Now we have to calculate the moles of conjugate base or salt $(CH_3NH_3Cl)$.

Using Henderson Hesselbach equation :

$pOH=pK_b+\log \frac{[Salt]}{[Base]}$

$pOH=pK_b+\log \frac{\text{Moles of Salt}}{\text{Moles of Base}}$

Now put all the given values in this expression, we get:

$4.3=3.36+\log (\frac{\text{Moles of Salt}}{0.1})$

$\text{Moles of Salt}=0.871mol$

Therefore, the moles of weak base $(CH_3NH_2)$ needed to added is, 0.871 mol.