Question

You carry out the titration of 25.00 mL of 0.250 M Nitrous Acid (HNO2) using a 0.375 M NaOH solution

Nitrous acid Ka= 4.5*10^-4

Calculate the pH at the volume of NaOH added, see table

Titration point	Volume of NaOH added (mL)	pH
1	0
2	4.25
3	8.33
4	16.67

a. At 0 mL NaOH

b. At 4.25 mL NaOH

c. At 8.33 mL NaOH

d. at 16.67 mL NaOH

Answer 1

millimoles of HNO2 = 25 x 0.250 = 6.25

pKa = -log (4.5 x 10^-4) = 3.35

a) 0.0 mL NaOH :

pH = 1/2 (pKa - log C)

    = 1/2 (3.35 - log 0.250)

pH = 1.98

b) 4.25 mL NaOH :

millimoles of NaoH = 4.25 x 0.375 = 1.594

HNO2   +    NaOH    -------------->   NaNO2   +   H2O

6.25          1.594                               0                0

4.656             0                                1.594

pH = pKa + log [salt / acid]

    = 3.35 + log [1.594 / 4.656]

pH = 2.88

c) At 8.33 mL NaOH

millimoles of NaoH = 8.33 x 0.375 = 3.124

this is half - equivalence point :

here

pH = pKa

pH = 3.35

d) at 16.67 mL NaOH :

millimoles of NaOH = 16.67 x 0.375 = 6.25

this is equivalence point , here salt remains.

salt concentration = 6.25 / (16.67 + 25) = 0.15

pH = 7 + 1/2 (pKa + log C)

    = 7 + 1/2 (3.35 + log 0.15)

pH = 8.26