Question

1. A 100.0-mL sample of 0.500 M sodium hydroxide is titrated with 0.100 M nitric acid. Calculate the pH after 400.0 mL of acid has been added.

2. A 1.0-L buffer solution contains 0.100 mol HCN and 0.100 mol LiCN. The value of Ka for HCN is 4.9 x 10^-10. Because the initial amounts of acid and conjugate base are equal, the pH of the buffer is equal to pKa = -log (4.9 x 10^-10) = 9.31. Calculate the new pH after the addition of 0.010 mol of HNO₃ to the buffer. Ignore any change of volume brought about by this addition.

3. Calculate the pH of a 1.0-L solution that is 0.50 M in NH₃ and 0.20 M in NH₄Cl after the addition of 30 mL of 1.0 M HNO₃ to the buffer. pKb = 4.75.

Answer 1

Sol 1.

Millimoles of nitric acid , HNO3

= concentration of HNO3 × volume of HNO3

= 0.1 M × 400 mL = 40 mmol  

Millimoles of sodium hydroxide , NaOH

= concentration of NaOH × volume of NaOH

= 0.5 M × 100 mL = 50 mmol

Reaction : NaOH + HNO3 ----> NaNO3 + H2O

So , millimoles of NaOH remain after neutralization reaction

= millimoles of NaOH - millimoles of HNO3

= 50 mmol - 40 mmol = 10 mmol  

Total volume of solution = 100 mL + 400 mL = 500 mL

Concentration of NaOH remain after neutralization reaction

= [NaOH] = 10 mmol / 500 mL = 0.02 M

So , [OH-] = [NaOH] = 0.02 M

pOH = - log[OH-] = - log(0.02) = 1.70

Therefore , pH = 14 - pOH

= 14 - 1.70  

= 12.30 ( answer )