1) when 10.0mL was added:
first moles of NH3 before adding the 10mL HCl = 0.3 m * 0.055 L = 0.0165 M
then moles of HCl after adding the 10mL = 0.5 m * 0.01 L = 0.005 M
when 1 mol HCl → 1 mol NH4+
So, 0.005 M HCl→ 0.005 M NH4+
∴ moles of NH4+ after adding 10 mL = 0.005 M
moles of NH3 after adding 10mL = 0.0165M - 0.005M = 0.0115 M
When Kb for NH3 = 1.82x10^-5
So, accordig to the reaction equation:
NH3 + H2O ↔ NH4+ + OH-
from the equation so, Kb formula is:
Kb= [OH-][NH4+]/[NH3]
so by substitution we can get the [OH] value:
1.82x10^-5 = [OH-]* 0.005 / 0.0115
∴[OH] = 4.2x10^-5
∴POH = -㏒[OH]
= -㏒(4.2x10^-5)
= 4.38
∴PH = 14- POH
= 14 - 4.38 = 9.62
2)when 20mL was added:
When we have moles of NH3 before adding the 20mL of HCl = 0.0165M
∴moles HCl after adding the 20mL = 0.5 m * 0.02L =0.01 M
∴ moles NH4+ after adding the 20mL = 0.01 M
∴moles of NH3 after adding 20mL = 0.0165 M - 0.01 M
= 0.0065 M
when we have:
NH3 + H2O ↔ NH4+ + OH-
and Kb of NH3 = 1.8x10^-5
when Kb = [OH-][NH4]/[NH3]
by substitution:
1.8x10^-5 = [OH-]* 0.01 / 0.0065
∴[OH-] = 1.17x10^-5
∴POH = -㏒(1.17x10^-5)
= 4.93
∴PH = 14 - POH
= 14 - 4.93
= 9.1
3) when 51.5 mL was added:
moles NH4 before adding 51.5 mL of HCl = 0.0165 M
moles HCl after adding 51.5 mL of HCl = 0.5 m * 0.0515 L= 0.02575 M
∴moles NH4 after adding 51.5 mL of HCl in 106.5mL (the total volume)=
=0.0165 / total volume
= 0.0165 / 0.1065 = 0.155
NH4+ → H+ + NH3
assume that [NH3] = [H+] = X
when Ka = [H+][NH3] /[NH4+]
so by substitution:
1.8x10^-5 = X^2 / 0.155
X = 0.00167
∴[H+] = 0.00167
∴PH = - ㏒[H+]
= - ㏒ 0.00167
= 2.78
