Calculate the pH of a buffered solution after adding strong acid [closed]

Question

Calculate the pH of a solution formed by the addition of $0.0020$ moles of $\ce{HCl}$ to $\pu{100.0 mL}$ of $\pu{0.200 M}$ $\ce{CH3NH2}$/ $\ce{0.100 M CH3NH3Cl}$.

$K_\mathrm{b}(\ce{CH3NH2})=4.4\times 10^{-4}$

Any hints will be much appreciated. :)

Answer 1

Okay, here goes nothing.

$$\ce{CH3NH3+ <=> H+ + CH3NH2}$$

$$K_a(\ce{CH3NH3+})=\frac{[\ce{H+}][\ce{CH3NH2}]}{[\ce{CH3NH3+}]}\tag{1}$$

$\displaystyle K_a(\ce{CH3NH3+})=\frac{10^{-14}}{K_b(\ce{CH3NH2})}=\frac{10^{-14}}{4.4\times 10^{-4}}=2.27\times 10^{-11}$

Initial moles of $\ce{CH3NH3+}=0.100\times 0.100=0.010$

Initial moles of $\ce{CH3NH2}=0.100\times 0.200=0.020$

$$\ce{H+ + CH3NH2 -> CH3NH3+}$$

$0.0020$ moles of $\ce{H+}$ react with $0.0020$ moles of $\ce{CH3NH2}$ to form $0.0020$ moles of $\ce{CH3NH3+}$.

Final moles of $\ce{CH3NH3+}=0.010+0.0020=0.012$

Final moles of $\ce{CH3NH2}=0.020-0.0020=0.018$

$V=100.0\text{ mL}=0.1000\text{ L}$

$\displaystyle[\ce{CH3NH3+}]_f=\frac{0.012}{0.1000}=0.120$

$\displaystyle[\ce{CH3NH2}]_f=\frac{0.018}{0.1000}=0.180$

Substituting these values in equation $(1)$, we get

$\displaystyle[\ce{H+}]=\frac{2.27\times 10^{-11}\times[\ce{CH3NH3+}]}{[\ce{CH3NH2}]}=\frac{2.27\times 10^{-11}\times 0.120}{0.180}=1.51\times 10^{-11}$

pH$=-\log[\ce{H+}]=10.82$

Does this look good?