Question

1. 50.0 ml of 0.10 M HNO₂ (Ka = 4.0 x 10^-4) are being titrated with 0.10 M NaOH. The pH after 25.0 ml of NaOH have been added is...?

2. If 25 ml of 0.75M HCl are added to 100 ml of 0.25M NaOH, what is the final pH?

thank you!

Answer 1

1)

Ka of HNO2 = 4.0*10^-4
pka = -log Ka
= -log (4.0*10^-4)
= 3.4

mol of HNO2 present initially = M*V = 0.10 M * 50 mL = 5 mmol
mol of NaOH added = 0.10 M * 25 mL = 2.5 mmol

2.5 mmol of both will react to form 2.5 mmol of NaNO2
after reaction,
mol of NaNO2 = 2.5 mmol
mol of HNO2 remaning = (5-2.5) mmol = 2.5 mmol

so,
[NaNO2]/[HNO2] = 1

pH = pKa + log {[NaNO2]/[HNO2]}
= 3.4 + log (1)
= 3.4 + 0
= 3.4
Answer: 3.4

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