For the titration of 55.0 ml of 0.300 m nh3 with 0.500 m hcl at 25 c, determine the relative ph at each of these points.

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

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