Question

A 100.0 mL sample of 0.10 M NH3 is titrated with 0.10 M HNO3. Determine the pH of the solution after the addition of 100.0 mL of HNO3. The Kb of NH3 is 1.8 × 10-5.

Please show work. a) 10.56 b) 5.28 c) 6.58 d) 8.72 e) 3.44

Answer 1

we know that

moles = molarity x volume (L)

so

moles of NH3 taken = 0.1 x 100 x 10-3 = 10 x 10-3

moles of HN03 added = 0.1 x 100 x 10-3 = 10 x 10-3

the reaction is

NH3 + HN03 ---> NH4NO3

we can see that

moles of NH3 reacted = moles of HN03 added = 10 x 10-3

so

moles of NH4NO3 formed = moles of NH3 reacted = 10 x 10-3

now

total volume = 100 + 100 = 200 ml

now

concentration = moles x 1000 / volume (ml)

so

[NH4NO3]= 10 x 10-3 x 1000 / 200

[NH4N03] = 0.05

now

NH4N03 is a weak acid

we know that

for weak acids

[H+] = sqrt ( Ka x C)

also

Ka = 10-14 / Kb

so

[H+] = sqrt ( 10-14 x 0.05 / 1.8 x 10-5)

[H+] = 5.27 x 10-6

now

pH = -log [H+]

so

pH = -log 5.27 x 10-6

pH = 5.278

so

the answer is b) 5.28