Buffer solution

"A frequently used example is a mixture of ammonia solution and ammonium chloride solution. If these were mixed in equal molar proportions, the solution would have a pH of 9.25. It doesn't matter what concentrations you choose as long as they are the same." Why does not it matter what concentrations we choose as long as they are the same ?

#Chemistry

Easy Math Editor

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Comments

You should refer to read Buffer system from any good physical chemistry book of your choice.The reason why this is the case is that for a weak base-salt of weak base buffer system the pH is given by

$\text{pH} = \text{pK}_a + \log \dfrac{\text{Base}}{\text{Salt}}$

this is known as the Henderson-Hasselbalch equation. As you can see when the concentrations of base and salt are same, the logarithm term is equal to $\log(1) =0$ and hence $\text{pH} = \text{pK}_a$ . The value of $\text{pK}_a$ for ammonia is constant at the given temperature. Most problems give the value of ${pK}_b$ for ammonia but you can write in terms of ${pK}_a$ as ${pK}_a = {pK}_w - {pK}_b$ . The equation above in terms of ${pK}_b$ is

$\text{pOH} = \text{pK}_b + \log \dfrac{\text{Salt}}{\text{Base}}$

either way, the result is the same.

Tapas Mazumdar - 3 years, 2 months ago

It's pH or pOH

A Former Brilliant Member - 3 years, 2 months ago

Yes it is pOH if pKb is taken into equation. It should be pKa and also note that it is Base/Salt not Salt/Base. Thanks for pointing that out though. Silly mistake!

Tapas Mazumdar - 3 years, 2 months ago

@Tapas Mazumdar – You are welcome

A Former Brilliant Member - 3 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$