Question

What is the pH of a solution that results when 0.010 mol HNO3 is added to 500 mL of a solution that is 0.10 M in aqueous ammonia and 0.20 M in ammonium nitrate

Answer 1

Here is the full question.

What is the pH of a solution that results when 0.010 mol HNO3 is added to 500 mL of a solution that is 0.10 M in aqueous ammonia and 0.20 M in ammonium nitrate? Assume no volume change. The Kb of ammonia is 1.8*10^-5.

Answer:

8.82

Step-by-step explanation:

Volume = 500 mL = 0.500 L

Number of moles of NH₃ = 0.10 mole × 0.500 L = 0.050 moles

Number of moles of NH₄⁺ = 0.20 mole × 0.500 L = 0.10 moles

NH₃ + H⁺ ----------> NH₄⁺

In 0.010 mole of HNO₃ ;

Number of moles of NH₃ = 0.050 moles - 0.010 moles

= 0.040 moles

Number of moles of NH₄⁺ = 0.10 moles + 0.010 = 0.11 moles

Concentration of NH₃ =
`(number of moles)/(volume)`

=
`(0.040)/(0.500)`

= 0.080 M

Concentration of NH₃ =
`(number of moles)/(volume)`

=
`(0.11)/(0.50)`

= 0.220 M

NH₃ + H₂O ⇄ NH₄⁺ + OH⁻

Initial 0.080 M 0 0.220 M 0

Change -x +x x

Equilibrium 0.080 -x 0.220 +x x

`K_b` =
`((0.220+x)(x))/((0.080-x))`

`1.8*10^(-5)` =
`((0.220+x)(x))/((0.080-x))`

x = [OH⁻] = 6.55 × 10⁻⁶ M

pOH = 5.18

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 5.18

pH = 8.82