0
$\begingroup$

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?

$\endgroup$
9
  • 1
    $\begingroup$ 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). $\endgroup$
    – Poutnik
    Commented Aug 20, 2022 at 14:49
  • $\begingroup$ So what type of buffer would NH3 and NH4Cl create? $\endgroup$ Commented Aug 20, 2022 at 14:52
  • $\begingroup$ basic, as it is weak base + its salt with strong acid.// acidic buffer is weak acid + its salt with strong base. $\endgroup$
    – Poutnik
    Commented Aug 20, 2022 at 14:57
  • $\begingroup$ alright ill deit my original $\endgroup$ Commented Aug 20, 2022 at 15:00
  • $\begingroup$ Can you help me by answering the original question tho? $\endgroup$ Commented Aug 20, 2022 at 15:00

1 Answer 1

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.