Question

This is a two-part question if someone could please help!

110.0 −mL buffer solution is 0.110 M in NH3and 0.130 M in NH4Br.

Part A: What mass of HCl can this buffer neutralize before the pH falls below 9.00?

Part B: If the same volume of the buffer were 0.255 M in NH3  and 0.395 M in NH4Br, what mass of HCl could be handled before the pH falls below 9.00?

Answer 1

Initial number of NH₃ = Molarity x vol in lts = 0.11 M x 0.11 lts = 0.0121 moles

Initial number of NH₄⁺ = Molarity x vol in lts = 0.13 M X 0.11 lts = 0.0143 moles

Kb value of ammonia buffer = 1.8 x 10^-5

pKb = -log Kb = - log (1.8 x 10^-5) = 4.74

Let us consider X is Moles of HCl

Equilibrium conc. of NH₃ = 0.0121 - X (decreases its conc. after addition of HCl)

Equilibrium conc. of NH₄⁺ = 0.0143 + X (increases its conc after addition of HCl)

According to Henderson- Hasselbalch equation :

pOH = pKb + log [ NH₄⁺]/[NH₃]

pOH = 14-pH = 14 - 9 = 5

5 = 4.74 + log (0.0143 + X/0.0121-X)

5-4.74 = log (0.0143 + X/0.0121-X)

log (0.0143 + X/0.0121-X) = 0.26

0.0143 + X/0.0121-X) = Antilog (0.26)

0.0143 + X/0.0121-X) = 1.82

0.0143 + X = 1.82 (0.0121- X)

0.0143 + X = 0.022 - 1.82 X

0.0143 - 0.022 = - 1.82 X - X

-0.0077 = -2.82 X

X = 0.00273 moles of HCl

Molar mass of HCl = 36.46 g/mol

Mass of HCl = Moles x molar mass of HCl = 0.00273 moles x 36.46 g/mol = 0.0995 g

B) Initial moles of NH₃ = Molarity x vol in lts = 0.255 M x 0.11 lts = 0.02805 moles

Initial number of NH₄⁺ = Molarity x vol in lts = 0.395 M X 0.11 lts = 0.04345 moles

Kb value of ammonia buffer = 1.8 x 10^-5

pKb = -log Kb = - log (1.8 x 10^-5) = 4.74

Let us consider X is Moles of HCl

Equilibrium conc. of NH₃ = 0.02805 - X (decreases its conc. after addition of HCl)

Equilibrium conc. of NH₄⁺ = 0.04345 + X (increases its conc after addition of HCl)

According to Henderson- Hasselbalch equation :

pOH = pKb + log [ NH₄⁺]/[NH₃]

pOH = 14-pH = 14 - 9 = 5

5 = 4.74 + log (0.04345 + X/0.02805 - X)

5-4.74 = log (0.04345 + X/0.02805 - X)

log (00.04345 + X/0.02805 - X) = 0.26

0.04345 + X/0.02805 - X) = Antilog (0.26)

0.04345 + X/0.02805 - X) = 1.82

0.04345 + X = 1.82 (0.02805 - X)

0.04345 + X = 0.0511 - 1.82 X

0.04345- 0.0511 = - 1.82 X - X

-0.00765 = -2.82 X

X = 0.00271 moles of HCl

Molar mass of HCl = 36.46 g/mol

Mass of HCl = Moles x molar mass of HCl = 0.00271 moles x 36.46 g/mol = 0.0988 g