Answer to Question #154374 in Chemistry for Amnah

Question

At certain conditions, 11.5 g of nitrous acid is dissolved in 2.0 L of water. The pH is 3.36. What is the Ka for these conditions? 

Answer 1

According to the equation:

Ka = [H+][A-] / [HA] = [H+] [NO₂-] / [HNO₂]

where [H+] - protons concentration, [A-] - anion concentration, [HA] - acid concentration.

As 11.5 g of nitrous acid is dissolved in 2.0 L of water:

M (HNO₂) = m / (Mr × V) = 11.5 g / (47.013 g/mol × 2.0 L) = 0.12 M

As pH = 3.36:

[H+] = 10^-pH = 10^-3.36 = 0.0004 M

As nitrous acid is a strong acid:

Ka = [0.00004] [0.00004] / [0.12 - 0.00004] = 1.3 × 10^-8 M

Answer: 1.3 × 10^-8 M