These are hydrolysis problems.
.......HCO2^- + HOH --> HCOOH + OH^-
I......0.037.............0......0
C........-x..............x......x
E......0.037-x...........x.......x
Kb(for HCO2^-) = (Kw/Ka for HCOOH) = (HCOOH)(OH^-/(HCO2^-)
Substitute from the E line and solve for x = (OH^-). Then I would convert this to pOH = -log(OH^-) and to pH by
pH + pOH = pKw = 14.
You can obtain (H^+) from
pH = -log(H^+).
Calculate the hydronium ion concentration and pH in a 0.037 M solution of sodium formate, NaHCO2.
hydronium ion concentration
i don't understand these
2 answers
i did what you told me to and had to use the quadratic and got 1.430e-6 for the OH- but it was wrong. is that what you got?