Question

A 130.0 −mL buffer solution is 0.105 M in NH3 and 0.135 M in NH4BrNH4Br. The Kb value for NH3 is 1.76×10−5

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 0.265 M in NH3  and 0.390 M in NH4Br, what mass of HCl could be handled before the pH falls below 9.00? Express the mass in grams to three significant figures.

Answer 1

Part A : mass of HCl = 0.0855 g

Part B : mass of HCl = 0.131 g

Explanation

Part A

concentration of NH3 = 0.105 M

volume of buffer = 130 mL = 0.130 L

initial moles of NH3 = (concentration of NH3) * (volume of buffer in Liter)

initial moles of NH3 = (0.105 M) * (0.130 L)

initial moles of NH3 = 0.01365 mol

Similarly, initial moles of NH4+ = 0.01755 mol

Let moles of HCl added = x

HCl is a strong acid which will neutralize NH3 to NH4+

final moles of NH3 = (initial moles of NH3) - (moles of HCl added)

final moles of NH3 = 0.01365 mol - x

final moles of NH4+ = (initial moles of NH4+) + (moles of HCl added)

final moles of NH4+ = 0.01755 mol + x

pH = 9.00

pOH = 14 - pH

pOH = 14 - 9.00

pOH = 5.00

Kb = 1.76 x 10-5

pKb = -log(Kb)

pKb = -log(1.76 x 10-5)

pKb = 4.75

According to Henderson - Hasselbalch equation,

pOH = pKb + log([conjugate acid] / [weak base])

pOH = pKb + log(final moles of NH4+ / final moles of NH3)

5.00 = 4.75 + log[(0.01755 mol + x) / (0.01365 mol - x)]

log[(0.01755 mol + x) / (0.01365 mol - x)] = 5.00 - 4.74

log[(0.01755 mol + x) / (0.01365 mol - x)] = 0.26

(0.01755 mol + x) / (0.01365 mol - x) = 100.26

(0.01755 mol + x) / (0.01365 mol - x) = 1.76

solving for x, x = 0.00235 mol

moles of HCl added = x = 0.00235 mol

mass of HCl added = (moles of HCl added) * (molar mass HCl)

mass of HCl added = (0.00235 mol) * (36.46 g/mol)

mass of HCl added = 0.0855 g