Question

To 1 litre solution containing 0.1 mole each of NH3 and NH4Cl, 0.05 mole of NaOH is added. The change in pH will be (pkb for NH3=4.74)

Answer 1

Answer:- Change in pH is 0.48.

Solution:- We have a buffer solution since the weak base(ammonia) and it's salt or conjugate acid(ammonium chloride) are present in the solution. The pH is easily calculated using Handerson equation:

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

pKa could be calculated from given pKb as:

Pka = 14 - pKb

Pka = 14 - 4.74 = 9.26

let's plug in the values in the equation to calculate the original pH of the buffer:

$pH=9.26+log(\frac{0.1}{0.1})$

pH = 9.26 + 0

pH = 9.26

Now, 0.05 moles of NaOH are added. It's a strong base so it would react with the acid(ammonium ion) present in the buffer as:

$NH_4^+(aq)+OH^-(aq)\rightleftharpoons NH_3(aq)+H_2O(l)$

(sodium and chloride ions are not included here as they are the spectator ions and so they get canceled out.)

ammonium ion and hyroxide ion(NaOH) reacts in 1:1 mol ratio to give ammonia(base). So, 0.05 moles of hydroxide ion will react with exactly 0.05 moles of ammonium ion to form 0.05 moles of ammonia.

So, new moles of $NH_4^+$ = 0.1 - 0.05 = 0.05

new moles of $NH_3$ = 0.1 + 0.05 = 0.15

Let's plug in the values in the Handerson equation again:

$pH=9.26+log(\frac{0.15}{0.05})$

pH = 9.26 + 0.48

pH = 9.74

Change in pH = 9.74 - 9.26 = 0.48

So, the change in pH on addition of 0.05 moles of NaOH to the given buffer solution is 0.48.