$$HNO_2 \rightleftharpoons H^+ + NO_2^-$$
The pH of a weak acid can be calculated using the following formula:
$$pH = -\log[H^+]$$
where [H^+] is the hydrogen ion concentration in moles per liter (M).
The dissociation constant (Ka) for nitrous acid is 4.5 x 10^(-4) at 25°C. The Ka is a measure of the strength of an acid, and the smaller the Ka, the weaker the acid. For nitrous acid:
$$Ka = [H^+][NO_2^-]/[HNO_2]$$
Assuming that x is the concentration of H^+ and NO2- ions produced at equilibrium, and the initial concentration of HNO2 is C, then:
$$[H^+] = [NO_2^-] = x$$
$$[HNO_2] = C - x$$
Substituting these concentrations into the Ka expression:
$$4.5 \times 10^{−4} = x^2/(C - x)$$
At equilibrium, the concentration of the conjugate base, NO2-, is small compared to the initial concentration of HNO2, so we can assume that C ≈ [HNO2] in the denominator. Therefore, simplifying the equation, we have:
$$x^2 + (4.5 \times 10^{-4})x - (4.5 \times 10^{-4})C = 0$$
Solving for x, the hydrogen ion concentration:
$$x = \frac{-b ± √(b^2 - 4ac)}{2a}$$
where a = 1, b = 4.5 x 10^(-4), and c = -(4.5 x 10^(-4))C.
Calculating the hydrogen ion concentration (x):
$$x = \frac{-(4.5 \times 10^{-4}) ± √((4.5 \times 10^{-4})^2 - 4(1)(-4.5 \times 10^{-4})C)}{2(1)}$$
$$x = \frac{4.5 \times 10^{-4} ± 0.0198C}{2}$$
Since the hydrogen ion concentration cannot be negative, we take the positive root:
$$x = \frac{0.0198C + 4.5 \times 10^{-4}}{2}$$
Substituting the Ka expression into the equation:
$$x = \frac{Ka[HNO_2] + Ka}{2}$$
$$x = \frac{(4.5 \times 10^{-4})[HNO_2] + 4.5 \times 10^{-4}}{2}$$
At 25°C:
$$pH = -\log \left(\frac{(4.5 \times 10^{-4})[HNO_2] + 4.5 \times 10^{-4}}{2}\right)$$
For example:
If [HNO2] = 0.1 M:
$$pH = -\log \left(\frac{(4.5 \times 10^{-4})(0.1) + 4.5 \times 10^{-4}}{2}\right) = 2.85$$
Therefore, the pH of a 0.1 M nitrous acid solution is approximately 2.85.
