Creating a buffer with with a target pH and given weak base?

I need to create a $100\textrm{mL}$ buffer of $\textrm{pH = 9.20}$ with ammonia and ammonium chloride such that $\textrm{pH = }9.20\pm0.50$ with $20\textrm{mL}$ of $0.2\textrm{M } \ce{NaOH}/\ce{HCl}$. I am provided with $0.1\textrm{M}$ ammonia and ammonium chloride salt.


I used the Henderson-Hasselbalch equation to figure out the relative amounts of $\ce{NH3}$ and $\ce{NH4+}$ I need:


$$
\textrm{pH} = \textrm{pKa} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
\textrm{9.20} = 9.25 + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
\frac{[\ce{NH3}]}{[\ce{NH4+}]} = 0.89
$$ 


So for every mole of $\ce{NH4+}$ I need $0.89$ moles of $\ce{NH3}$ to reach the desired pH of 9.20. I know that the buffer must withstand $0.004$ moles of $\ce{NaOH}/\ce{HCl}$, so I can set up the inequality


$$
8.70 < \textrm{pKa} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} < 9.70 \\
-0.55 < \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} < 0.45 \\
0.28 < \frac{[\ce{NH3}]}{[\ce{NH4+}]} < 2.82
$$


So I know the relative ratios I need to be within after the strong base and acid, but from here, how do I proceed with my $0.1\textrm{M}$ ammonia solution and ammonium chloride to get the desired ratio and buffer capacity? 




Your initial calculations using Henderson-Hasselbalch equation is correct:
$$
\mathrm{pH} = \mathrm{p}K\_\mathrm{a} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
\mathrm{9.20} = 9.25 + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
\log \frac{[\ce{NH3}]}{[\ce{NH4+}]} = -0.05\\
\frac{[\ce{NH3}]}{[\ce{NH4+}]} = 0.89
$$ 


Yet, since you have only $\pu{0.10 M}$ ammonia solution, you can't exceed $\ce{[NH3]}$ by $\pu{0.10 M}$. Thus, to prepare $\pu{100 mL}$ of ammonia buffer with $\mathrm{pH}=9.20$, you have to add $m$ grams of $\ce{NH4Cl}$ to $\pu{100 mL}$ of $\pu{0.10 M}$ ammonia solution where:


$$ m = \frac{\pu{0.10 molL-1}}{0.89}\times \pu{0.10 L}\times \pu{53.49 gmol-1}=\pu{0.601 g}$$ 


Assuming that the volume has not changed when you add $\pu{0.601 g}$ of $\ce{NH4Cl}$ to $\pu{100 mL}$ of ammonia solution, you have gotten the buffer you needed with $\mathrm{pH}=9.20$. Now, you have to check the buffer capacity by adding $\pu{20 mL}$ of either $\pu{0.20 M}$ solutions of $\ce{NaOH}$ or $\ce{HCl}$. 


Let's add $\pu{20 mL}$ of $\pu{0.20 M}$ solution of $\ce{NaOH}$ first. The following reaction would take place:
$$\ce{NH4Cl + NaOH -> NH3 + NaCl + H2O}$$


Thus, $0.20 \times \pu{0.020 mol}= \pu{0.004 mol}$ of $\ce{NH4Cl}$ will convert to $\pu{0.004 mol}$ of $\ce{NH3}$. The new volume of solution is $\pu{(100+20) mL}=\pu{120 mL}$.


The new $\ce{[NH3]} = \dfrac{\pu{(0.1\times 0.1+0.004) mol}}{\pu{0.120 L}}=\pu{0.117 molL-1}$


Similarly, The new $\ce{[NH4Cl]} = \dfrac{\pu{(\frac{0.1\times 0.1}{0.89} -0.004) mol}}{\pu{0.120 L}}=\pu{0.060 molL-1}$.


Therefore, the new pH would be:


$$
\mathrm{pH} = \mathrm{p}K\_\mathrm{a} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
= 9.25 + \log \frac{0.117}{0.060} = 9.54 \lt 9.70
$$ 


Thus, this buffer stands for the higher $\mathrm{pH}$ limit required. You can also check the lower limit by similar method. 




---


**Late edition:**


My friend MaxW insists in elsewhere that this is the case of $\ce{NH4Cl}$ solution to use the buffer (opposes to what I did with solid). Max correctly calculated volumes of two $\pu{0.10 M}$ solutions would be $\pu{52.91 mL}$ of $\ce{NH4Cl}$ and $\pu{47.09 mL}$ of $\ce{NH3}$. If Max was correct, this should still stand the required buffer capacity. Let's check:


The new $\ce{[NH3]} = \dfrac{\pu{(0.1\times 0.04709 + 0.004) mol}}{\pu{0.120 L}}=\pu{0.0726 molL-1}$


Similarly, The new $\ce{[NH4Cl]} = \dfrac{\pu{(0.1\times 0.05291-0.004) mol}}{\pu{0.120 L}}=\pu{0.0108 molL-1}$.


Therefore, the new pH would be:


$$
\mathrm{pH} = \mathrm{p}K\_\mathrm{a} + \log \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\
= 9.25 + \log \frac{0.0726}{0.0108} = 10.08 \gt 9.70
$$ 


Thus, this buffer would not stand for the higher $\mathrm{pH}$ limit required.


 1

The question asks:



> 
> I need to create a $100\textrm{mL}$ buffer of $\textrm{pH = 9.20}$ with ammonia and ammonium chloride such that $\textrm{pH = }9.20\pm0.50$ with $20\textrm{mL}$ of $0.2\textrm{M } \ce{NaOH}/\ce{HCl}$. I am provided with $0.1\textrm{M}$ ammonia and ammonium chloride.
> 
> 
> 


~~User Mathew Mahindaratne has provided an answer, but did not answer the question. The question supposes that 0.1 molar solutions of ammonia and ammonium chloride are to be used. Mathew used ammonium chloride salt.~~ 


**EDIT User Mathew Mahindaratne did in fact answer the question correctly. The OP edited the question and confirmed that a salt of $\ce{NH4Cl}$ was being used.**


To me there is an added constraint.


$$\mathrm{V\_\ce{NH3}} + \mathrm{V\_\ce{NH4Cl}} = \pu{100 ml}\tag{1}$$


From Mathew's answer


$$\dfrac{[\ce{NH3}]}{[\ce{NH4+}]} = 0.89\tag{2}$$


If we neglect the amount of $\ce{NH4OH}$ in the ammonia solution, then the concentrations of ammonia and ammonium are equal so:


$$\dfrac{\mathrm{V\_\ce{NH3}}}{\mathrm{V\_\ce{NH4+}}} = 0.89\tag{3}$$


Substituting into equation 1 yields:


$$
0.89\*\mathrm{V\_\ce{NH4Cl}} + \mathrm{V\_\ce{NH4Cl}} = \pu{100 ml} \\
\mathrm{V\_\ce{NH4Cl}} = \dfrac{\pu{100 ml}}{1.89} = \pu{52.91 ml}\tag{4} 
$$
So $\pu{47.09 ml}$ of the ammonia solutioon must also be used. 


 1