Question

What is the pH of a solution that results when 0.012 mol HNO3 is added to 610.0 mL of a solution that is 0.26 M in aqueous ammonia and 0.46M in ammonium nitrate. Assume no volume change. (Kb for NH3 = 1.8 × 10–5.)

a. 8.96

b. 9.56

c. 5.04

d. 9.01

e. 4.44

Answer 1

[NH3] = molarity x volume in Litres = 0.26 M x 0.61 L = 0.1586 mol

[NH4NO3] = molarity x volume in Litres = 0.46 M x 0.61 L = 0.2806 mol

[HNO3] = 0.012 mol

     NH3 +      HNO3      ------------------> NH4NO3

0.1586 mol     0.012 mol                      0

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0.1586-0.012           0                                0.012 mol

= 0.1466 mol

Hence,

[NH3] = 0.1466 mol

[NH4NO3] = 0.012 mol + 0.2806 mol = 0.2926 mol

Given that Kb for NH3 = 1.8 x10^-5

We know that Henderson-Hasselbalch equation is

pOH = - logKb + log [NH4NO3]/[NH3]

       = - log (1.8 x10^-5) + log (0.2926 mol/ 0.1466 mol)

       = 5.04

pOH = 5.04

Hence,

pH = 14 - pOH = 14 - 5.04 = 8.96

Therefore,

pH = 8.96

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