The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
2007-03-02 07:52:56 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
NaHCO2 is a strong electrolyte so we get 0.35 M of HCO2-
HCO2- + H2O>> HCOOH +OH-
Kh= HCOOH OH-/HCO2- = Kw/Kdiss=5.5 10^-11
Let x=moles/L HCOO2- that hydrolyze .
This will give us x moles/L HCOOH and x moles/L OH- while redicing the concentration of HCO2- from 0.35 to 0.35-x
5.5 10^-11= (x)(x)/0.35-x
x=4.38 10^-6=OH-
pOH=5.36
pH=8.64
2007-03-02 08:07:29 · answer #1 · answered by Anonymous · 1⤊ 0⤋
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pH of an aqueous solution?
The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?
2015-08-16 11:26:22 · answer #2 · answered by Anonymous · 0⤊ 0⤋