# Answered Essay: Formic acid, HCHO_2, has a K_a = 1.8 times 10^-4. A student is asked to prepare a buffer

Formic acid, HCHO_2, has a K_a = 1.8 times 10^-4. A student is asked to prepare a buffer having a pH of 3.40 from a 0.10 M HCHO_2 and a 0.10 M NaCHO_2 (sodium formate: formate ion is the conjugate base). How many milliliters of the NaCHO_2 solution should she add to 20.0 mL of the 0.10 M HCHO_2 to make the buffer? When 5 drops of 0.10 M NaOH were added to 20 mL of the buffer in problem 3, the pH went from 3.40 to 3.43. Write a net ionic equation to explain why the pH didn’t go up to about 10, as it would have if that amount of NaOH were added to distilled water or to 20.0 mL of 0.00040 M HCl, which also would have a pH of 3.40

3.
a. no of mole of HCOOH = 0.1*20/1000 = 0.002 mole

no of mole of HCOONa =

pH of acidic buffer = pka + log(salt/acid)

pka of HCOOH = 3.8

3.4 = 3.8+log(x/(0.002-x))

x = no of mole of HCOONa = 0.00057 mol

volume of HCOONa must be taken = n/M = 0.00057/0.1 = 0.0057 L

= 5.7 ML

4. addition of base (OH-) to acidic buffer

HCOOH(aq) + OH^-(aq) —-> HCOO-(aq) + H2O(l) results no change in pH

HCOO-(aq) + H3O^+(aq) —-> HCOOH(aq) + H2O(l) results no change in pH

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