pH of a solution from M and Ka

Question

What is the pH of a 0.320 M solution of Ca(NO₂)₂ (Ka of HNO₂ is 4.5 × 10⁻⁴)?

What I did :

0.320 x 4.5

x²= 1.44x 10^-4

x = 0.037947

-log (0.037947)

pH = 1.42

The software says that is incorrect, so I'm not sure what I'm doing wrong.

Answer 1

1st) You assume all of the calcium nitrite dissolves: M_Nitrite= 2 (.32M) = .64 M

2nd) You want to find the K_b of NO₂^- = K_w/K_A(HNO₂) = 10^-14/(4.5x10^-4)

That is the equation NO₂^- + H₂O → OH^- + HNO₂

3rd) Solve for [OH^-] = x K_b = x²/(.64) where water has an activity of 1, we assume x<<.64 (denominator is actually .64 -x and you are welcome to solve the quadratic equation)

4th) You have [OH^-] : you can find pOH and subtract from 14 to find pH or you can divide 10^-14 by [OH^-] and take the -log to find pH.

Good luck!