Question

To 1.0 L solution containing 0.1 mol each of NH, and NH CI, 0.05 mol NaOH is added. The change in pH will be (pKa of CH,COOH=4.74)

Answer 1

• 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.

pH = 9.26 + 0.48

pH = 9.74

Change in pH 

= 9.74 - 9.26 

= 0.48

Comments

I confuse. What is difference between hydrogen and hydronium ion respect to ionic change.will you explain