Question

A 110.0 −mL buffer solution is 0.110 M in NH3 and 0.135 M in NH4Br.

Part A What mass of HCl could this buffer neutralize before the pH fell below 9.00? Express your answer using two significant figures.

m=?

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 fell below 9.00? Express your answer using two significant figures.

m=?

Answer 1

pKb of NH₃= 4.744727495
Initial number of moles of NH₃
= ([NH₃])(volume of buffer solution in L)
= (0.110)(110x10⁻³)
= 0.0121

Initial number of moles of NH₄⁺
= ([NH₄⁺])(volume of buffer solution in L)
= (0.135)(110x10⁻³)
= 0.01485

pOH= pKb+log([NH₄⁺]/[NH₃])
= 4.744727495+log(0.135/0.110)
= 4.853871964 + 0.089 = 4.9428

pH= 14.0 - 4.9428
= 9.0572

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

5.00-4.9428= log([NH₄⁺]/[NH₃])
[NH₄⁺]/[NH₃]= 10^(5.00-4.49428)
= 3.2
Number of moles of NH₄⁺/number of moles of NH₃= 3.2

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

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

Number of moles of NH₃ remaining after the reaction
= 0.0121-x
Number of moles of NH₄⁺ present after the reaction
= (initial number of moles of NH₄⁺)+(number of moles of HCl added)
= 0.01485+x

(0.01485+x) / (0.0121-x)= 3.2
0.01485 + x = 0.03872 -3.2x
4.2x= 0.02387

x = 5.68 X 10^-3
Relative Formula Mass(RFM) of HCl
= (RAM of H)+(RAM of Cl)
= 1+35.5
= 36.5

1 mole of HCl= 36.5g
2.507142857x10⁻³ moles of HCl
= (36.5)(5.68x10⁻³)
= 0.207g correct to 3 sf

Thus, if more than 0.207 g of HCl were added to this buffer solution, its pH would fall below 9.00. Part (B) is to be approached the exact same way, except that you change the values of the initial [NH₃] and [NH₄⁺] to 0.265M and 0.390M respectively.