Question

To 25.00 mL of 0.100 M NH₃ solution are added 15.00 mL of 0.100 M HCl solution. (K_b NH₃ = 1.8 x 10^-5)

By how much will the pH change if a further 15.00 mL of the 0.100 M HCl solution are added?

• A.

  2.72

• B. 5.61

• C.

  8.32

• D.

  7.04

• E.

  6.39

Answer 1

pKb of NH3 = - log (1.8x10^-5) ) = 4.75

part A)

NH3 + HCl -----------------------> NH4Cl

25x0.1 =2.5 15x0.1=1.5 0 initial mmoles

1.0 0 1.5

So pH of this buff er is calculated using Hendersen equation as

pOH = pKb + log[conjugate acid] /[base]

= 4.75 + log 1.5/1.0

=4.926

and

pH = 14-pOH = 14-4.926 =9.074

part B)

NOw another 15 mL of acid added

NH3 + HCl -----------------------> NH4Cl

25x0.1 =2.5 30x0.1=3.0 0 initial mmoles

0 0 .5 2.5

the solution now has excess of strong acid HCl.

thus the [H+] in solution = mmoles / total volume = 0.5/55 =0.00909 M

and pH = -log [H+] = -log (0.00909) =2.04

Thus the chang ein pH = previous pH - pH afer adding HCl

=9.074 - 2.04

= 7.034

OPTION D