Question

A 110.0 mL buffer solution is 0.100 M in NH3 and 0.125 M in NH4Br. (Kb of NH3 is 1.76×10−5.)

question: If the same volume of the buffer were 0.250 M in NH3  and 0.395 M in NH4Br, what mass of HCl could be handled before the pH falls below 9.00?

Help ASAP Please!!!!!

Answer 1

Answer:

Starting number of moles of NH?

= ([NH?])(volume of cushion arrangement in L)

= (0.250)(110x10-³)

= 0.0275

Starting number of moles of NH??

= ([NH??])(volume of cushion arrangement in L)

= (0.395)(110x10?³)

= 0.04345

Kb of NH3 is 1.76×10-5

pKb=-log(1.76×10-5)=4.754487

pOH= pKb+lg([NH??]/[NH?])

= 4.754487+log(0.395/0.250)

= 4.953

pH= 14-4.953

= 9.046856

Be that as it may, the pH is 9.00, So pOH=5

pOH-pKb=log([NH??]/[NH?])

5.00-4.754487= lg([NH??]/[NH?])

[NH??]/[NH?]= 10^(5.00-4.754487)

= 1.76

Number of moles of NH??/number of moles of NH?= 1.76

At the point when HCl is included, the accompanying response happens:

NH?(aq)+HCl(aq)?NH?Cl(aq)

As per the condition, the proportion of the quantity of moles of NH? that responds to that of HCl that responds is 1:1. Give x a chance to be the quantity of moles of HCl included

NH?(aq)+HCl(aq)?NH?Cl(aq)

starting: 0.0275 0.04345

change: - x +x

balance: 0.0275-x 0.04345+x

Presently, (0.04345+x)/(0.0275-x)= 1.76

2.76x= -0.3861

x= -0.13989

1 mole of HCl= 36.5g

-0.13989 moles of HCl

= (36.5)(-0.13989)

= -5.105985 g