In a titration of 100 mL 0.10M nitrous acid HNO2 against 0.10M sodium hydroxide, calculate the pH at the following stages:
K(a) for the nitrous acid = 4.5 x 10-4

a) before any base has been added to reach the acid solution.

b) after 25% of the base necessary to reach the equivalence point has been added.

c) at the equivalence point.

In a titration of 100 mL 0.10M nitrous acid HNO2 against 0.10M sodium hydroxide, calculate the pH at the following stages:
K(a) for the nitrous acid = 4.5 x 10-4

a) before any base has been added to reach the acid solution.


a. Calculate the concentration of H+

HNO_2 \rightleftharpoons H^+ + NO_2^-

Take the negative log of that concentration to find the pH.

0.1M HNO2. You immediately know the concentration.

\frac {[H^+][NO_2^-]}{[HNO_2]}=4.5 \times 10^{-4} (Equation 1)

\frac {x^2}{0.1-x}=4.5 \times 10^{-4} where x is the amount that's dissociated.

x^2+4.5 \times 10^{-4}x - 4.5 \times 10^{-5}=0

x= \frac {-4.5 \times 10^{-4} \pm \sqrt {( 2.025 \times 10^{-9} + 1.8 \times 10^{-4} )}}{2} \approx \frac {-4.5 \times 10^{-4} \pm 1.34 \times 10^{-2}}{2}

the negative root makes no chemical sense so

x \approx 1.30 \times 10^{-2}

x is both the hydrogen ion concentration and the nitrite ion concentration.

pH = -log(1.30 \times 10^{-2}) \approx 1.89



b) after 25% of the base necessary to reach the equivalence point has been added.


b. Add the base. Figure out the concentration of NO2 -- don't forget to add the volume of base that's added. Assume that the volumes add. Then, do the same calculation as in #1.



In a titration of 100 mL 0.10M nitrous acid HNO2 against 0.10M sodium hydroxide, calculate the pH at the following stages:
K(a) for the nitrous acid = 4.5 x 10-4

c) at the equivalence point.


c. Here the amount of acid equals the amount of base. You have only sodium nitrite (NaNO2). Nitrous acid is a weak acid:

NO_2^- + H_2O \rightleftharpoons HNO_2 + OH^-

This is really not all that different from Equation 1 since [H^+][OH^-]=10^{-14}

Figure out the concentration of Nitrite. Create an equation like I did in problem #1. In this case, the conc of HNO2 equals the conc of OH-

\frac {(10^{-14} - [OH^-])[NO_2^-]}{[HNO_2]}=4.5 \times 10^{-4} (Equation 1b)

if C is the concentration of HNO2 before dissociation

\frac {(10^{-14} - x)(C-x)}{x}=4.5 \times 10^{-4} (Equation 1b)

Solve for x and calculate the pH.