A method to create NH3-NH4Cl buffer solution using titration method

My aim is to create a basic buffer solution.


I know that to create a basic buffer solution, we need a strong acid and weak base. In my case I have chosen $\ce{NH3}$ and $\ce{NH4Cl}$.


Using titration, I first want to create $\ce{NH4Cl}$ shown in equation below.


$$\ce{NH3 + HCl -> NH4Cl}$$


After this step I am a bit confused.


Should I mix the $\ce{NH4Cl}$ and $\ce{NH3}$ to create the buffer?


Also I want to create a buffer solution with $\mathrm{pH} = 10$. Can anyone tell me the concentration and volumes required for each of the chemicals in this case?




As Poutnik correctly pointed out in the conmment section, titration is just a small detail in preparation of buffer solution. Actually it was a little confution to me. Yet, titriometric calculations can be used to set the conjugate acid/base pair ratio for a buffer with perticular $\mathrm{pH}$. Still, the $\mathrm{pH}$ value of the buffer is approximate value and cannot be predicted exactly. The exact value can be achieved by using a $\mathrm{pH}$ meter and either a dilute sulution of $\ce{NaOH}$ (if $\mathrm{pH}$ is below the expected value) or a dilute sulution of $\ce{HCl}$ (if the $\mathrm{pH}$ is above the expected value).


There are few methods to prepare buffer solutions. For example, you can simply calculate the weak base (e.g., $\ce{NH3}$) and its conjugated acid (e.g., $\ce{NH4Cl}$) using Henderson-Hasselbalch equation for the given $\mathrm{pH}$. Suppose you need to prepare $\ce{NH3/NH4Cl}$ buffer at $\mathrm{pH} = 10.0$. The relevant Henderson-Hasselbalch equation is:


$$\mathrm{pOH} = \mathrm{p}K\_\mathrm{b} + \log \left(\frac{[\ce{NH4+}]}{[\ce{NH3}]}\right)$$


Since you know the $\mathrm{pOH}$ $\left(= 14.0 -\mathrm{pH} = 14.0 - 10.0 = 4.0\right)$ and $\mathrm{p}K\_\mathrm{b}$ of $\ce{NH3}$ ($4.75$), you can calculate the ratio of $\left(\frac{[\ce{NH4+}]}{[\ce{NH3}]}\right)$:


$$\left(\frac{[\ce{NH4+}]}{[\ce{NH3}]}\right) = 10^\left(\mathrm{pOH} - \mathrm{p}K\_\mathrm{b}\right) = 10^\left(4.0 - 4.75\right) = 1.78$$


If you have $\pu{0.1 M}$ $\ce{NH3}$ solution, you need to have $\pu{0.1 \times 1.78 M}$ $\ce{NH4+}$ ($\pu{0.178 M}$) in you buffer solution. Assuming volume of the $\ce{NH3}$ solution woud not increse when certain amount of solid is dissoved, you can dissolve $0.178 \times \pu{53.5 g} = \pu{9.52 g}$ of solid $\ce{NH4Cl}$ in $\pu{1.0 L}$ of $\pu{0.1 M}$ $\ce{NH3}$ solution. That would give you the requred buffer solution with $\mathrm{pH}$ around $10$. Put a $\mathrm{pH}$ meter in the solution to measure accurrate reading. If that value is above $10$, use $\pu{3 M}$ $\ce{HCl}$ solution to readjust solution to $\mathrm{pH} = 10.0$. Similarly, if the value is below $10$, use $\pu{3 M}$ $\ce{NaOH}$ solution to readjust it to $\mathrm{pH} = 10.0$.


Now, you need to prepare a $\ce{NH3/NH4+}$ buffer at $\mathrm{pH} = 10.0$ using $\ce{NH3}$ solution and a $\ce{HCl}$ solution without using solid $\ce{NH4Cl}$.


Suppose you have conc. $\ce{HCl}$ and $\pu{500 mL}$ $\pu{0.2 M}$ $\ce{NH3}$ solution to prepare sought buffer solution. We need higher concentration of $\ce{NH3}$ solution beecause part of $\ce{NH3}$ is going to convert to $\ce{NH4+}$ during the process after strong acid/weak base reaction:


$$\ce{NH3 + HCl -> NH4+ + Cl-}$$


The conc. $\ce{HCl}$ solution should be used to minimize the increment of the volume significantly. We know by previous calculations that after adding enough conc. $\ce{HCl}$ ($\approx \pu{15 mL}$), the ratio of $\left(\frac{[\ce{NH4+}]}{[\ce{NH3}]}\right)$ should still be $1.78$ if the $\mathrm{pH}$ reading of $\mathrm{pH}$ meter indicates $\approx 10$. Thus, we should put a $\mathrm{pH}$ meter in $\pu{500 mL}$ $\pu{0.2 M}$ $\ce{NH3}$ solution to measure accurrate reading when we add conc. $\ce{HCl}$ dropwise (in this way, we don't need to measure exact amount of conc. $\ce{HCl}$). When the $\mathrm{pH}$ meter reads exactly $10.0$, stop addition of conc. $\ce{HCl}$ and dilute the buffer solution to $\pu{1.0 L}$ using deionized water. At this point, the $\mathrm{pH}$ meter reading would change, but you can readjust it to $10.0$ as follows:


If the $\mathrm{pH}$ of diluted solution is above $10$, use $\pu{3 M}$ $\ce{HCl}$ solution to readjust solution to $\mathrm{pH} = 10.0$. Similarly, if the value is below $10$, use $\pu{3 M}$ $\ce{NaOH}$ solution to readjust it to $\mathrm{pH} = 10.0$.


Note that $[\ce{NH3}]$ in each solution prepared by method 1 and method 2 are not identical as well as $[\ce{NH4+}]$ in those solutions. However, $\left(\frac{[\ce{NH4+}]}{[\ce{NH3}]}\right)$ is still $1.78$. If you need accurate concentrations of each species (e.g., $[\ce{NH3}] + [\ce{NH4+}]$), you should do more accurate calculations.


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