Calculate the hydronium ion concentration and the pH at the

Calculate the hydronium ion concentration and the pH at the equivalence point when 55.0 mL of 0.7000 M NH₃ is mixed with 65.0 mL of 0.5923 M HCl.
K_a=5.6x10^-10
[H₃O⁺]
M  
pH

Solution

NH3 + HCl = NH4Cl

1 mole NH3 reacts with 1 mole HCl

Number of moles NH3 = 0.055*0.7 = 0.0385

Number of moles HCl = 0.065*0.5923 = 0.0385

Thus what you have is a solution of 0.0385 moles NH4Cl in total volume 120 mL

NH4+ will associate with water to produce some H+ ions because NH3 is a weak base and NH4+ is the conjugate acid.

NH4+ + H2O = NH3 + H3O+

Ka = [NH3][H+]/[NH4Cl]

we can use the number of moles because volume is constant

Suppose x moles of NH4+ dissociate

then [H+] = [NH3] = x and [NH4Cl] = 0.009-x

Assume x << 0.009

then Ka = x*x/0.009 = 5.6*10^-10

x^2 = 0.009*5.6*10^-10
= 5.04*10^-12

x = 2.245*10^-6 = [H+]

pH = 5.65