Question

A 100.0 −mL buffer solution is 0.110 M in NH3 and 0.125 M in NH4Br.

A. If the same volume of the buffer were 0.260 M in NH3 and 0.395 M in NH4Br, what mass of HCl could be handled before the pH fell below 9.00?

B. What mass of HCl could this buffer neutralize before the pH fell below 9.00?

Answer 1

A. pOH = pKb+ log{ [NH4+]/[NH3]}

NH4Br supliments NH4+

pKb for NH3= 4.74

pOH= 4.74+ log [ 0.395/0.260) =4.92

pH=14-pOH= 14-4.92= 9.08

for the case when pH fell below 9,

poH= 14-9= 5

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

[NH4+]/[NH3] =1.82

HCl addition influences the following reaction NH3 +HCl ----> NH4Cl

so there will be more of NH4+ from NH4Cl and less of NH3 diue to its consumption

let x= moles of HCl added,

moles of NH3 initially = 0.11*100/1000 =0.011 and moles of NH4Br= 0.125*100/1000 =0.0125

let x= moles of HCl added

hence moles of NH4+ =0.0125+x and moles of NH3= 0.011-x

(0.0125+x)/ (0.011-x) =1.82

0.0125+x =1.82*0.011-1,82x

2.82x= 0.00752

x = 0.00772/2.82=0.002667

mass of HCl to be added = 0.002667*36.5 =0.097 gm

2. for the 1st case, pOH= 4.74 +log (0.125/0.11)=4.795

pH= 14-4.795=9.205

when the pH falls below 9,

pOH= 14-9 =5

pOH= pKb+log[ NH4+]/[NH3]

[NH4+]/[NH3] =1.8

let x= moles of HCl added

then NH4+ =0.395*100/1000 +x =0.0395+xand NH3= 0.260*100/1000-x =0.026-x

0.0395+x =1.8*(0.026-x)

2.8x= 1.8*0.026-0.0395,x =0.002607, mass of HCl = 0.002607*36.5 =0.095 gm

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