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

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 = $\frac{number of moles}{volume}$

= $\frac{0.040}{0.500}$

= 0.080 M

Concentration of NH₃ = $\frac{number of moles}{volume}$

= $\frac{0.11}{0.50}$

= 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

$K_b$ =  $\frac{(0.220+x)(x)}{(0.080-x)}$

$1.8*10^{-5}$ =  $\frac{(0.220+x)(x)}{(0.080-x)}$

x = [OH⁻] = 6.55 × 10⁻⁶ M

pOH = 5.18

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 5.18

pH = 8.82