# 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?

• I know that to create a basic buffer solution, we need a strong base and weak acid. - Exactly the opposite. This is acidic buffer (e.g. acetic acid+ sodium hydroxide). Aug 20, 2022 at 14:49
• So what type of buffer would NH3 and NH4Cl create? Aug 20, 2022 at 14:52
• basic, as it is weak base + its salt with strong acid.// acidic buffer is weak acid + its salt with strong base. Aug 20, 2022 at 14:57
• alright ill deit my original Aug 20, 2022 at 15:00
• Can you help me by answering the original question tho? Aug 20, 2022 at 15:00

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.