What is the pH of a solution prepared by dissolving 1.5g HCOONa in 250mL of solution

mols HCOONa = g/molar mass = 1.5/68 = about 0.022

M HCOONa = moles/L = 0.022 moles/0.25 L = about 0.088 M
......................HCOO^- + HOH --> HCOOH + OH^-
I..........................0.088......................0..............0
C.........................-x............................x..............x
E.....................0.088-x........................x..............x
Kb for HCOO^- = Kw/Ka for HCOOH = (HCOOH)(OH^-)/(HCOO^-)
Kw/Ka = (x)(x)/(0.088-x)
You know Kw, you can look up Ka, solve for x = (OH^-)
Then pOH = -log(OH^-). Finally,
pH + pOH = pKw = 14 to solve for pH.
Post your work if you get stuck.