Question

To prepare a buffer solution of PH 8.26, amount of (NH4)2SO4 to be added into 500ml of 0.01M NH4OH solution if pkb(NH4+) is 9.26

Answer 1

Answer:- $1.995*10^-^4g$

Solution:- The pH of a buffer solution is calculated by using the Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

For the given buffer, the weak base is ammonium hydroxide and the acid is ammonium sulfate as it is a salt of weak base(ammonium hydroxide) and strong acid(sulfuric acid).

moles of ammonium hydroxide can be calculated from its given volume and molarity as:

$500mL(\frac{1L}{1000mL})(\frac{0.01mol}{L})$

= 0.005 mol

pKa is calculated from given pKb using the formula:

pKa = 14 - pKb

pKa = 14 - 9.26

pKa = 4.74

pH is given as 8.26, Let's plug in the given values in handerson equation and calculate the moles of acid that is ammonium sulfate.

$8.26=4.74+log(\frac{0.005}{n})$

$8.26-4.74=log(\frac{0.005}{n})$

$3.52=log(\frac{0.005}{n})$

On taking antilog:

$10^3^.^5^2=log(\frac{0.005}{n})$

$3311.31=(\frac{0.005}{n})$

$n=\frac{0.005}{3311.31}$

$n=1.51*10^-^6$

Molar mass of $(NH_4)_2SO_4$ is 132.14 gram per mol.

Let's multiply the moles by molar mass to get the mass in grams.

$1.51*10^-^6mol(\frac{132.14g}{mol})$

= $1.995*10^-^4g$

So, $1.995*10^-^4g$ of ammonium sulfate are required to prepare the buffer.