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

Which of the following did you identify correctly

Answer : option c) heat is released during the process.explanation :

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.