Question

A.) What is the pH of a solution that contains 50.00 mL of 0.10 M NH3 to which 60.00 mL of 0.10 M HCl has been added? (please include equation, table and explain)

B.) Calculate the pH of 0.25 M NH4Cl

Answer 1

we know that

moles = concentration x volume (L)

so

moles of NH3 = 0.1 x 50 x 10-3 = 5 x 10-3

moles of HCl = 0.1 x 60 x 10-3 = 6 x 10-3

now

the reaction is

HCl + NH3 ---> NH4Cl

we can see that

moles of HCl reacted = moles of NH3 present = 5 x 10-3

so

moles of HCl remaining = 6 x 10-3 - 5 x 10-3 = 1 x 10-3

also

moles of NH4Cl formed = moles of NH3 present = 5 x 10-3

now

there are 2 species in the solution

NH4Cl and HCl

we know that

HCl is a very strong acid

so

[H+] mostly comes from HCl

now

final volume = 50 + 60 = 110 ml

so

[HCl] = 1 x 10-3 / 110 x 10-3 = 9.0909 x 10-3

HCl --> H+ + Cl-

we can see that

[H+] = [HCl]

so

[H+] = 9.0909 x 10-3

now

pH = -log [H+]

so

pH = -log 9.0909 x 10-3

pH = 2.04

so

pH of the solution is 2.04

B)

given 0.25 M NH4Cl

we know that

NH4+ + H20 --> NH3 + H30+

Ka = [NH3] [H30+] / [NH4+]

Ka = [x] [x] / [0.25-x]

Ka = x2/ [0.25-x]

we know that

Ka for NH4+ is 5.623 * 10-10

5.623 * 10-10 = x2 / [0.25-x]

x = 1.1856 * 10-5

now

pH = -log [H30+]

so

pH = -log 1.1856 * 10-5

pH = 4.926

so

the pH of the solution is 4.926