NaNO2 is the salt of a strong base (NaOH) and a weak acid (HNO2). The NO2^- is hydrolyzed.
NO2^- + HOH ==> HNO2 + OH^-
Kb = (Kw/Ka) = (HNO2)(OH^-)/(NO2^-)
Set up an ICE chart.
You know Kw. You can calculate Ka from pKa. (HNO2) will be x. (OH^-) will be x. (NO2^-) = 0.262. Solve for x, convert to pOH, then to pH. I found pH = 8.285 which I would round to 8.28 but I would pick the 8.29 for the answer.
I GOT THIS WRONG IN AN ASSIGNMENT. I HAVE AN EXAM IN 2 DAYS. CAN YOU SHOW ME HOW TO DO IT PLS.
What is the pH of a 2.62 x 10 -1 M NaNO2 solution ? NB the pKa for HNO2 = 3.153.
12.13
13.42
1.87
8.29
0.58
1 answer
1 answer