Question

calculate the pH of a solution obtained by mixing 500 ml of 0.02M HCl with 500 ml of 0.04M NH3. Given: pKb for NH3 = 4.74

Answer 1

Answer:As in the above reacn. You can see there is a formation of basic buffer so,After reacn. 0.01molar NH4Cl and 0.1 molar NH3 is left.

So, it is the best basic buffer so,

POH=PKb

So, PH=14–4.74=9.26

I hope it helps you.☺

Explanation:

Answer 2

The pH of a solution obtained by mixing 500 ml of 0.02M HCl with 500 ml of 0.04M NH3 is :

Given: pKb for NH3 = 4.74

• No. of moles of NH3 = 500×10^-3×0.04 = 0.02 moles

• No. of moles of HCl = 500×10^-3×0.02 = 0.01 moles

• Here we have limiting agent as HCl.

Reaction : HCl + NH3 => NH4+ + Cl-

• We have 1:1 mole ratio for above reaction.

• After reaction, moles of HCl would be zero as it completely gets consumed.

• Moles of NH3 after reaction = 0.02 - 0.01 = 0.01 moles

• Mixture made up is of weak base and its conjugate acid i.e Buffer reaction.

• Total volume = 500 ml + 500 ml = 1000 ml = 1 L

• Concentration of NH4+ and NH3 would be

[NH4+] = [NH3] = 0.01 moles /1 L = 0.01 M

pKb = 4.74

• pOH = pKb + log [NH4+]/[NH3]

[NH4+] = [NH3]

• Hence log [NH4+]/[NH3] = 0

pOH = pKb

pOH = 4.74

• pH = 14 - pOH = 14 - 4.74 = 9.26