Answer to Question #180944 in General Chemistry for Patricia

Question

You add 100.0 mL of 0.50 M NaOH to 50.0 mL of 1.00 M NH4Cl. What is the pH of the resulting solution? Kb(NH3) = 1.8 x 10^-5. Provide answer with solution.

Answer 1

Moles of NaOH= c x V= 0.50x100/1000= 0.050mol

Mole of NH4CL= CxV= 1.00 x 50/1000= 0.050mol

The two compounds dissociate as follows

NaOH---> Na+ + OH-

Number of moles of OH- produced = 0.050mol

Total volume of solution= 100.0ml + 50.0ml= 150.0ml = 0.15L

Concentration of OH-= 0.050mol/0.15L= 0.33M

Also,

NH4Cl----> NH4+ + Cl-

NH4+ undergoes hydrolysis

NH4+ + H20 ::::::: NH3 + H30+

Ka(NH4+) = 10^-14/Kb(NH3) = 10^-14/1.8x10^-5 = 5.6x10^-10

Concentration of NH4+= 0.050mol/0.15L = 0.33M

Equilibrium concentrations

[NH4+] = 0.33-x

[NH3]= x

[H3O+]= x

Ka= [NH3][H3O+]/[NH4+]

Ka= x²/0.33-x

Since NH4+ is weak, 0.33-x is approximately 0.33

x²= Ka x 0.33

x²= 5.6x10^-10 x 0.33

x²= 1.848x10^-10

x= 1.36x10^-5M

[H30+]= 1.36 x 10^-5M

Now,

[OH-] - [H30+] = 0.33M - (1.36x10^-5)

which is approximately = 0.33M of OH-

pOH= -log(OH-)

pOH= -log(0.33)

pOH= 0.48

pH= 14-0.48

PH= 13.52