Question

A 130.0 mL buffer solution is 0.110 M in NH3 and 0.135 M in NH4Br. (Kb of NH3 is 1.76×10−5.)

If the same volume of the buffer were 0.255 M in NH3 and 0.400 M in NH4Br, what mass of HCl could be handled before the pH falls below 9.00?

Please help me figure out the answer! I am on my last attempt. I have entered the following answers and these are NOT correct: 0.0916, 0.0915, 0.094, 0.101, 0.999, 0.0999, 2.24

Answer 1

Initial number of moles of NH₃
= ([NH₃])(volume of buffer solution in L)
= (0.110)(130x10⁻³)
= 0.0143

Initial number of moles of NH₄⁺
= ([NH₄⁺])(volume of buffer solution in L)
= (0.135)(130x10⁻³)
= 0.01755

pOH= pKb+lg([NH₄⁺]/[NH₃])
= 4.75448+lg(0.135/0.110)
= 4.843421

pH= 14-4.843421
= 9.1565789

When the pH is 9.00, pOH
= 14-9.00
= 5.00

5.00-4.75448= lg([NH₄⁺]/[NH₃])
[NH₄⁺]/[NH₃]= 10^(5.00-4.75448)
= 1.76
Number of moles of NH₄⁺/number of moles of NH₃= 1.76

When HCl is added, the following reaction occurs:
NH₃(aq)+HCl(aq)→NH₄Cl(aq)

According   equation, the ratio of number of moles of NH₃ that reacts to that of HCl that reacts is 1:1. Let a be the number of moles of HCl added.

Number of moles of NH₃ remaining after the reaction
= 0.0143 - a
Number of moles of NH₄⁺ present after reaction
= (initial number of moles of NH₄⁺)+(number of moles of HCl added)
= 0.01755 + a

(0.01755+a)/(0.0143 - a)= 1.76
2.76x= 7.6x10⁻³
x= 2.76014 x 10⁻³

Molecular mass of HCl
= 1+35.5
= 36.5

1 mole of HCl= 36.5g
2.76014 x 10⁻³ moles of HCl
= (36.5)(2.76014 x 10⁻³ )
= 0.100745 g Ans