Question

Calculate the pH at 25°C of 176.0 mL of a buffer solution that is 0.210 M NH4Cl and 0.210 M NH3 before and after the addition of 1.50 mL of 6.0 M HNO3. (The pKa for NH4+ = 9.75)

Answer 1

Ans :-

Explanation :-

1). Given that ,The pK_a for NH₄⁺ = 9.75

So pK_b = 14- pKa = 14-9.75= 4.25

From Henderson-Hasselbalch equation, we have

pOH = pK_b + log [salt / acid ]

so

pOH = pK_b + log [NH₄Cl / NH₃]

also

Here, pK_b for ammonia is 4.25

so

pOH = 4.25 + log [ 0.210 / 0.210]

pOH = 4.25

Now

pH = 14 - pOH

so

pH = 14 - 4.25

pH = 9.75

2). 1.50 mL of 6.0 M HNO₃.

HNO₃ is added

Moles of HNO₃ added = 6.00 x 1.50 x 10^-3 = 9.0 x 10^-3 mol

moles of NH₄Cl = 0.210 x 176 x 10-3 = 0.03696

moles of NH₃ = 0.210 x 176 x 10-3 = 0.03696

now

the reaction is given by

NH₃ + H⁺ ---> NH₄⁺

Now,

moles of NH₃ reacted = moles of HNO₃ added = 9.00 x 10^-3

moles of NH₄⁺ formed = moles of HNO₃ added = 9.00 x 10^-3

so

finally

moles of NH₃ = 0.03696 – 9.0 x 10^-3 = 0.02796

moles of NH₄⁺ = 0.03696 + 9.0 x 10^-3 = 0.04596

now

pOH= pK_b + log [NH₄⁺ / NH₃]

so

pOH = 4.25 + log [0.04596/0.02796]

pOH = 4.466

pH = 14 - pOH

so

pH = 14 – 4.466

pH = 9.53

Hence, pH after addition of HNO₃ is 9..53