pH = 8.7 so pOH = 14-8.7 = 5.3
5.3 = -log(OH^-) and (OH^-) = approx 5E-6 but you need to that more accurately.
........NO2^- + HOH ==> HNO2 + OH^-
I.......M................0......0
C......-x................x......x
E......M-x...............x......x
Kb for NO2^- = (Kw/Ka for HNO2) = (x)(x)/(M-x). You know x (that's OH^-) and Kw and you can look up Ka. Solve for M.
Suppose you wanted to produce an aqueous solution of pH = 8.70 by dissolving one of the following salts in water: NH4Cl, KHSO4, KNO2, NaNO3. Use salt KNO2.
But at what molarity?
Please help!! Thank you!
1 answer