Question

To 1 litre solution containing 0.1 mole each of NH3 and NH4Cl, 0.05 mol of NaOH is added. The change in pH will be ( pKb for NH3 = 4.74) ? Answer is 0.48

Answer 1

Why have you given answer?

Answer 2

Dear Friend,

Initial pH of the solution can be calculated as

[NH3]= No. of moles/Volume=0.1/1=0.1 M

Similarly, [NH4Cl] = 0.1 M

According to Henderson's equation

pOH = Kb + log[Salt][Base]

pOH = 4.74 + log0.1/0.1

or, pOH = 4.74

Now, pH = 14−pOH = 14−4.74 = 9.26

When 0.05 mole of NaOH is added NaOH + NH4Cl → NaCl + NH4OH

0.05 0.1 0 0.1 0 0.05 0.05 0.15

Now, pOH2 = 4.74 + log 0.05/0.15

pOH2 = 4.74 + log 0.3333

pOH2 = 4.74 − 0.4771 =4.26

pH2=14−4.26 = 9.74

Change in pH = pH1− pH2 = −0.48

Thus we can say that the pH of the solution increases by 0.48

Hope, it helps（＾＿－）≡★