Answer
-- pH of the buffer is 9.26
-- pH of the buffer is after addition of HCl is 9.18
Work Step by Step
-- Let's calculate [OH$^-$] at equilibrium to find the buffer's pH:
$NH_{3}$(aq) + $H_{2}$O (l) -->$NH_{4}^+$ (aq) + $OH^-$ (aq)
The equilibrium concentration of $NH_{4}^+$= (0.20+x)M
The equilibrium concentration of $OH^-$= 0+x =x M
The equilibrium concentration of $NH_{3}$= (0.20-x)M
K$_b$($NH_{3}$)= $\frac{[NH_{4}^+][OH^-]}{[NH_{3}]}$= $\frac{(0.20+x)(x)}{0.20-x}$=1.8$\times$10$^{-5}$
As the initial concentrations of $NH_{3}$ and $NH_{4}^+$, 0.20, is much greater than K$_b$($NH_{3}$), [$NH_{4}^+$]=0.20+x$\approx$0.20M and [$NH_{3}$]=0.20-x $\approx$0.20M.
So K$_b$($NH_{3}$)= $\frac{(0.20)(x)}{0.20}$=1.8$\times$10$^{-5}$ and [$OH^-$]=x=1.8$\times$10$^{-5}$M
pH=14-pOH=14-(-log[$OH^-$])=14+log(1.8$\times$10$^{-5}$)$\approx$9.26
-- To find the pH of the buffer after addition of HCl, let's follow the following steps:
Step 1: Find the initial concentrations of $NH_{4}^+$, $OH^-$ and $NH_{3}$ right after we add the 10.0 mL 0.10M HCl.
$NH_{3}$(aq) + HCl (aq) --> $NH_{4}^+$(aq) +$Cl^-$(aq)
Before addition of HCl, the buffer solution has 0.20M$\times$65.0mL=13mmol $NH_{3}$, (at equilibrium) and we add 0.10M$\times$10.0mL=1mmol HCl, so $NH_{3}$ is the excess reagent and HCl is the limit reagent.
The initial concentration of $NH_{3}$ right after adding HCl=$\frac{(13-1)mmol}{(65.0+10.0)mL}$=0.16M
Before addition of HCl, the buffer solution has 0.20M$\times$65.0mL=13mmol $NH_{4}^+$ and 1.8$\times10^{-5}$M$\times$65.0mL=1.17$\times10^{-3}$mmol $OH^-$ (both at equilibrium)
The initial concentration of $NH_{4}^+$ right after adding HCl=$\frac{(13+1)mmol}{(65.0+10.0)mL}$$\approx$0.19M
The initial concentration of $OH^-$ right after adding HCl=$\frac{1.17\times10^{-3}mmol}{(65.0+10.0)mL}$=1.56$\times10^{-5}$M
Step 2: Find [OH$^-$] at equilibrium to find the buffer's pH after addition of HCl.
As $NH_{3}$ reacts with HCl, so the equilibrium in $NH_{3}$(aq) + $H_{2}$O (l) -->$NH_{4}^+$ (aq) + $OH^-$ (aq) shifts to the left; therefore [$NH_{3}$] increases and both [$NH_{4}^+$] and [$OH^-$] decrease.
The equilibrium concentration of $NH_{4}^+$= (0.19-x)M
The equilibrium concentration of $OH^-$= (1.56$\times10^{-5}$-x)M
The equilibrium concentration of $NH_{3}$=(0.16+x) M
K$_b$($NH_{3}$)= $\frac{[NH_{4}^+][OH^-]}{[NH_{3}]}$= $\frac{(0.19-x)(1.56\times10^{-5}-x)}{0.16+x}$=1.8$\times$10$^{-5}$
As the initial concentrations of $NH_{3}$ and $NH_{4}^+$ are much greater than K$_b$($NH_{3}$), [$NH_{4}^+$]=0.19-x$\approx$0.19M and [$NH_{3}$]=0.16+x $\approx$0.16M.
So K$_b$($NH_{3}$)= $\frac{(0.19)(1.56\times10^{-5}-x)}{0.16}$=1.8$\times$10$^{-5}$ and [$OH^-$]=1.56$\times10^{-5}$-x$\approx$1.52$\times$10$^{-5}$M
pH=14-pOH=14-(-log[$OH^-$])=14+log(1.52$\times10^{-5})\approx$9.18