Question

(4). Chapter 14

Calculate the pH of the buffer prepared by mixing

(a)  2.0 g of NaOH with 12 g of NH₄Cl and diluting to 500 ml with wate

(b) 5.0 ml of 6.0 M HCl with 5.0 g of sodium acetate and diluting to 500 ml with water

Answer 1

a)
mol of NaOH = massof NaOH / molar mass of NaOH
= 2.0 g / 40 g/mol
= 0.05 mol

mol of NH4Cl = mass of NH4Cl/molar mass of NH4Cl
= 12 g / 53.5 g/mol
= 0.22 mol

0.05 mol of NH4+ will react with 0.05 mol of NaOH to form 0.05 mol of NH3
mol of NH4Cl remaining = 0.22 - 0.05 = 0.17 M

volume = 500 mL = 0.5 L

Kb of NH3 = 1.8*10^-5
pKb = -log Kb = -log (1.8*10^-5) = 4.745

use:
pOH = pKb + log {[NH4+]/[NH3]}
pOH = 4.745 + log (0.17/0.05)
= 5.28

pH = 14 - pOH
= 14 - 5.28
= 8.72

Answer: 8.72

b will be considered as different question as it is not related to a.

I am allowed to answer only 1 question at a time