Question

Be sure to answer all parts. Exactly 100 mL of 0.15 M nitrous acid (HNO2) are titrated with a 0.15 M NaOH solution. (a) Calculate the pH for the initial solution. (b) Calculate the pH for the point at which 80.0 mL of the base has been added. (c) Calculate the pH for the equivalence point. (d) Calculate the pH for the point at which 105 mL of the base has been added.

Answer 1

Molarity of nitrous acid (HNO2) = 0.15 M

Volume of HNO2 = 100 mL

Molarity of NaOH = 0.15 M

Ka of HNO2 = 4.5 * 10^–4   

pKa = - log (4.5 * 10^–4)

= 3.35

(a). pH for the initial solution.

Initially volume of NaOH added = 0 mL

                                      HNO₂ (aq)     H⁺ (aq)    +    NO₂^- (aq)
Initially                              0.15                       0                     0
Change                              - x                       +x                    +x
Final                                0.15 - x                   x                      x

Ka = x*x / (0.15 - x)

4.5 * 10^–4 = x*x / (0.15 - x)

x = 0.00799

[H+] = 0.00799 M

pH = - log [H+]

= - log (0.00799)

= 2.09

(b). pH for the point at which 80.0 mL of the base has been added.

Moles of HNO2 = 0.1 * 0.15 = 0.015

Moles of NaOH = 0.08 * 0.15 = 0.012

                       HNO2      +      OH-              NO2-      +        H2O
Initial               0.015              0.012                   0                      0
Change          - 0.012           - 0.012                0.012                0.012
Final                0.003                  0                    0.012               0.012

pH = pKa + log [salt] / [base]

= 3.35 + log (0.012/ 0.003)

= 3.95

(c). pH for the equivalence point.

At equivalence point, mmoles of NO2- formed = 0.015

Total volume = 100 + 100 = 200 mL

= 0.20 L

[NO2-] = 0.015 / 0.20

= 0.075 M

                        NO2- (aq) + H₂O (l)    HNO₂ (aq) + OH^- (aq)
Initial                 0.075            -                  0                  0
Change               - x               -                 + x               +x
Final               0.075 - x                              x                 x

Kb = x * x / (0.075 - x )

2.22 * 10^-11 = x * x / (0.075 - x )

x = 1.29 * 10^-6

[OH-] = 1.29 * 10^-6 M

pOH = - log (1.29 * 10^-6)

pOH = 5.89

pH = 14 - pOH

= 14 - 5.89

= 8.11

(d) pH for the point at which 105 mL of the base has been added.

Moles of OH- added = 0.105 * 0.15 = 0.01575

                       HNO2      +      OH-              NO2-      +        H2O
Initial               0.015             0.01575                 0                      0
Change          - 0.015           - 0.015                0.015                0.015
Final                0                0.00075             0.015                0.015

Total volume = 100 + 105 = 205 mL

= 0.205 L

[OH-] = 0.00075 / 0.205

= 0.003658 M

pOH = - log (0.003658)

= 2.43

pH = 14 - 2.43

= 11.57