Question

Weak Acid - Strong Base Titration

Calculate the pH at the given amounts below in the titration. You are titrating 35.0 mL of .15 M HNO₂ with .1 M NaOH until you reach 100 mL of NaOH added.

For this problem the Ka value is 4.6 x 10^-4

a) 0.00 mL NaOH added

b) 30.0 mL NaOH added

c) 60.0 mL NaOH added

d) 100.0 mL NaOH added

Answer 1

pKa of HNO2 = 3.34

a)

pH = 1/2 (pKa - log C)

     = 1/2 (3.34 - log 0.15)

pH = 2.09

b)

millimoles of HNO2 = 35 x 0.15 = 5.25

millimoles of NaOH = 30 x 0.1 = 3.0

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

5.25                 3.0                   0               0

2.25               0                            3.0

pH = pKa + log [salt / acid]

     = 3.34 + log [3.0 / 2.25]

pH = 3.46

c)

millimoles of NaoH = 60 x 0.1 = 6

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

5.25                 6.0                   0               0

   0               0.75                     5.25

here strong base remains.

[NaOH] = 0.75 / (60 + 35) = 7.89 x 10^-3 M

pOH = -log (7.89 x 10^-3 ) = 2.10

pH = 11.90

d)
millimoles of NaOH =100 x 0.1 = 10

[OH-] = 10 - 5.25 / (100 + 35) = 0.0352 M

pH = 12.55