Question

What mass of NH4Cl must be added to 250.0 mL of 0.200 M NH3 solution to have a buffer solution at pH=8.90? Assume no volume change.

A 15.0-mL sample of a H3PO4 solution is titrated with a 1.00 M NaOH solution. The neutralization reaction is complete when 33.6 mL of NaOH is added. What is the concentration of the H3PO4 solution (in M)?

Answer 1

Kb of NH3= 1.8*10-5

NH4Cl suppliments NH4+

NH3+H2O ------>NH4⁺+OH-

given pH= 8.9, pOH=14-8.9=5.1

[OH-]= 10^(-5.1)= 7.94*10^-6

Kb for the given reaction is Kb= [NH4+] [OH-]/[NH3]

1.8*10^-5 = [NH4+] 7.94*10^-6/0.2

[NH4+]= 1.8*10^-5*0.2/(7.94*10^-6)= 0.45M

1 L of solution contains 0.45 moles of NH4Cl

0.25L solution contains 0.45*0.25/1 moles of NH4Cl=0.1125

mass of NH4Cl= moles* molar mass= 0.1125*53.5= 6.02gm

2. The balanced reaction between H3PO4 and NaOH is

H3PO4+3NaOH-------->Na3PO4+3H2O

1 moles of H3PO4 requires 3 moles of NaOH for neutralization

moles of NaOH= molarity* volume in L= 1*33.6/1000 =0.0336

moles of H3PO4 required= 0.0336/3=0.0112

volume of H3PO4= 15ml= 15/1000 L=0.015

Concentration of H3PO4= moles/L=0.0112/0.015=0.75M