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…..

14)The Ka for formic acid (HCH02) is 1.8 X104. What is the pH of a 0.35-M aqueous solution of sodium formate (NaCH02)? a. 5.36 b. 10.71 C. 4.20 d: 8.64 e. 3.29

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

Expert Answer

Answer

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

In aqueous media CHO2 ion is in equilubria, HCOO + H2leftrightharpoons 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 + H2leftrightharpoons 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

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