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

Need 3 and 4 answered.

Formic acid, HCHO2, has a K. 1.8 X 104. A student is asked to prepare a buffer having a pH of 3.40 from a 0.10 M HCHO2 and a 0.10 M NaCHO2 (sodi conjugate base). How many mililiters of the NacHO2 solution should she add to 20.0 mL of 3. ium formate, formate ion is the the 0.10 M HCHO2 to make the buffer? mL 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 didnt 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 HCI, which also would have a pH of 3.40 4.

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

Expert Answer

Answer

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

   similarly if HCl added

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

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