# Question & Answer: The Ka for formic acid (HCHO_2) is 1.8 times 10^-4. What is the pH of a 0.35-M aqueous…..

The Ka for formic acid (HCHO_2) is 1.8 times 10^-4. What is the pH of a 0.35-M aqueous solution of sodium formate (NaCHO_2)? a. 5.36 b. 10.71 c. 4.20 d. 8.64 e. 3.29

K a ( formic acid ) = 1.8*10-4, concentration of NaCHO2 = .35M

In aqueous media CHO2 ion is in equilubria, HCOO + H2$\leftrightharpoons$ HCOOH + OH initial concentration of HCOOH and OH are 0,but later x amount of HCOOH and x amount of OH will form.

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HCOO + H2$\leftrightharpoons$ HCOOH +OH

.35 0 0

.35-x x x

then Kb is defined like, Kb = $\left [HCOOH] + \left [ OH^{-} ]$ $\left [HCOOH] + \left [ OH^{-} ]/\left [ HCO^{-}_{2} ]$

Kb = $x^{2}/\left [ HCO^{-}_{2} ]$

Ka*Kb = 10-14 , Kb = 10-14/1.8*10-4

Kb =5.56 *10-11 = $x^{2}/.35 -x$ , since $.35 > > x$ Kb = $x^{2^{}}$/.35

then $x$ =4.41*10-6 = $\left [OH^{-}]$ , pOH = -log $\left [OH^{-}]$   = 5.35

pH = 14 – pOH

= 8.6 , option d is correct