Question

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?

Two sig. figs.

m= ?

Answer 1

Which Volume are you talking about?

Assuming 130 mL,

pKb of NH₃= 4.744727495
Initial number of moles of NH₃
= ([NH₃])(volume of buffer solution in L)
= (0.265)(130x10⁻³)
= 0.03445

Initial number of moles of NH₄⁺
= ([NH₄⁺])(volume of buffer solution in L)
= (0.390)(130x10⁻³)
= 0.0507

pOH= pKb+lg([NH₄⁺]/[NH₃])
= 4.744+log(0.0507/0.03445)
= 4.912

pH= 14-4.912
= 9.088

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

5.00-4.744= lg([NH₄⁺]/[NH₃])
[NH₄⁺]/[NH₃]= 10^(5.00-4.744)
= 1.8
Number of moles of NH₄⁺/number of moles of NH₃= 1.8

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
= 0.03455-x
Number of moles of NH₄⁺ present after the reaction
= (initial number of moles of NH₄⁺)+(number of moles of HCl added)
= 0.0507+x

(0.0507+x)/(0.03445-x)= 1.8
solving for x:
x= 0.00404

Relative Formula Mass(RFM) of HCl
= (RAM of H)+(RAM of Cl)
= 1+35.5
= 36.5

1 mole of HCl= 36.5g
0.00404 moles of HCl
= (36.5)(0.00404)
= 0.1474g
= 0.15 g correct to 2 sf