Question

one litre of the buffer solution contains 0.1 mole each of nh4oh and nh4cl what will be the pH of the solution when 0.01 mole of the HCL is added to it pkb of nh4oh such is equals to 4.74​

Answer 1

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:

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 + 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:

(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 = 0.1 - 0.05 = 0.05

new moles of = 0.1 + 0.05 = 0.15

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

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.