Question

A 130.0 −mL buffer solution is 0.105 M in NH3 and 0.135 M in NH4Br.

A) What mass of HCl could this buffer neutralize before the pH fell below 9.00? ​answer to this was 0.088g

B) If the same volume of the buffer were 0.265 M in NH3 and 0.395 M in NH4Br, what mass of HCl could be handled before the pH fell below 9.00?

Answer 1

We know that pOH= pKb + log {[NH4+]/[NH3]= 4.74+log(0.135/0.105)=4.85

pH= 14-4.85= 9.15

When the pH at pH= 9 is

pOH= 14-9=5
pOH= pKb + log {[NH4+]/[NH3}

5 =4.74+log [NH4+]/[NH3]

[NH4+]/[NH3]= 1.82

It is also equal to moles of NH4+/ moles of NH3 = 1.82 (1)

When HCl is added, NH3 reacts with HCl
NH₃(aq)+HCl(aq)→NH₄Cl(aq)

NH3 is consumed and NH4+ is added to number of initial moles.

Moles of NH3 initially = molarity* Volume(L)= 0.105*130/1000 = 0.01365

Moles of NH4+( from NH4Br)= 0.135*130/1000 =0.01755
let x= moles of HCl reacted. According to the reaction, 1 mole of HCl is consumed per every moles of NH3.
Number of moles of NH₃ remaining after the reaction
= 0.01365-x
Number of moles of NH₄⁺ present after the reaction
= 0.01755+x

From Eq.1
(0.01755+x)/(0.01365-x)= 1.82

0.01755+x= 0.02484-1.82x

2.82x= 0.02484-0.01755 . x =0.002585

Molar mass of HCl = 36.5 , mass of HCl =0.002585*36.5= 0.094gm

2.

We know that pOH= pKb + log {[NH4+]/[NH3]= 4.74+log(0.135/0.105)=4.85

pH= 14-4.85= 9.15

When the pH at pH= 9 is

pOH= 14-9=5
pOH= pKb + log {[NH4+]/[NH3}

5 =4.74+log [NH4+]/[NH3]

[NH4+]/[NH3]= 1.82

It is also equal to moles of NH4+/ moles of NH3 = 1.82 (2)

When HCl is added, NH3 reacts with HCl
NH₃(aq)+HCl(aq)→NH₄Cl(aq)

NH3 is consumed and NH4+ is added to number of initial moles.

if the same volume of the buffer were 0.265 M in NH3 and 0.395 M in NH4Br, what mass of HCl could be handled before the pH fell below 9.00?

moles of NH3 in 130ml of 0.265= 0.265*130/1000=0.03445, moles of NH4+ =0.395*130/1000=0.05135, let x= moles of Hcl added,

hence NH3= 0.03445-x and NH4+ =0.05135+x

from Eq.2, 0.05135+x =1.82*(0.03445-x)

x=0.004 moles of HCl, mass of HCl = 36.5*0.004=0.146 gm