Question

Calculate the hydronium ion concentration and the pH at the equivalence point when 65.0 mL of 0.4000 M NH3 is mixed with 35.0 mL of 0.7429 M HCl.

Ka=5.6x10-10

Answer 1

no of moles of NH3 = 0.4M x 0.065 L = 0.026 mol

no of moles of HCl = 0.7429 M x 0.035L = 0.026 mol

lets see the balanced equation

NH3 + HCl -----> NH4Cl

from this it is clear that on emole of NH3 and one mole of HCl will combine and give the one mole of NH4Cl

according ly 0.26 mol of NH3 and 0.26 mole of HCl will combine and give the 0.26 mol of NH4Cl

total volume = 65+35 = 100 mL = 0.1L

concentration of NH4Cl = 0.26 / 0.1 = 2.6 M

now construct the ICE table

           NH4Cl + H2O <---> NH3 + H3O+   + Cl-

I         2.6                              0            0

C       -x                               +x           +x

E      2.6-x                             +x              +x

Ka = [H3O+][NH3] / [NH4Cl]

5.6 x 10^-10 = [x][x] / [2.6-x]

x² + x 5.6 * 10^-10 - 1.456 * 10^-9 = 0

solve the quadratic equation

x = 3.82 x 10^-5 which is the concentration of hydronium ion = [H3O+]

pH = -log(H3O+)

pH = -log(0.0000382)

pH = 4.42