What is the ph of 0.20 m nitrous acid at equilibrium (ka = 4.6*10-4) ?

Plato User, Its A) 2.01

#PlatoLivesMatter

Nitrous acid is HNO2. Its dissociation is HNO2 = H (+) + NO2 (-). The equilibrium constant, Ka = [H+][NO2-]/[HNO2] = > x^2 / (0.20 - x). Given that it is a weak acid (Ka = 4.6 * 10^ - 4), you can use a very good approxiamation: 0.20 >> x = > 0.20 - x = 0.20, and this permits you to solve the value of x more easily. = > 4.6 * 10^-4 = (x^2) / 0.20 = > x^2 = 0.20 * 4.6 * 10^ - 4 = 0.92 * 10^-4 = > x = 9.59 * 10^ - 3. Now, calculate pH = log { 1 / [H+] } = log { 1 / (9.59 * 10 ^-3) } = 2.02. Answer pH = 2.02