pH = 8.60
pOH = Kw/pH = about 6 but you need a better answer here and on the calculations below.
Convert pOH to (OH^-)
Let X = molarity of KNO2.
.........NO2%- + HOH ==> HNO2 + OH^-
I.........x...............0......0
C.........-y..............y......y
E.........x-y..............y.....y
Kb for NO2^- = Kw/Ka for HNO2 = (x)(x)/(x-y)
You know y from the OH^-. Substitute and solve for x. I'm guessing about 1 M.
Post your work if you get stuck.
Suppose you wanted to produce an aqueous solution of pH = 8.60 by dissolving one of the following salts in water:
KNO2
At what molarity?
1 answer