Question

Calculate the pH of a buffer which is 0.2moldm^-3 with respect to ammonium sulphate and 0.1 moldm^-3 with respect to ammonia. (Ka of NH4^+ = 5.6 x 10^-10 moldm^-3.

Answer 1

Answer:

Explanation:

Here, we want to get the pH of the buffer

To get this, we use the following equation:

$pH=pK_a\text{ + }log\text{ }\frac{\lbrack A^-\rbrack}{\lbrack HA\rbrack}$

Where:

$\text{pKa = -log(Ka)}$

We have the A^- as the concentration of the salt and HA is that of the weak acid

Now, let us calculate the pKa as follows:

$\text{pKa = -log (5.6 }\times10^{-10})\text{ = 9.25}$

From Let us get the concentration of the weak base and that of the salt:

For ammonium sulphate, we have:

$2\text{ moles of the ammonium ion from the sulphate}$

This means we have the concentration of the a