The formate buffer system contains formic acid (HCOOH) and sodium formate (NaCOO

Question

The formate buffer system contains formic acid (HCOOH) and sodium formate (NaCOOH). The pKa of formic acid is 3.74.

a) Write the equilibrium of this buffer system.

b) Calculate the concentration of formic acid and sodium formate to make a buffer that is pH 3.90. The total concentration of the buffer ([HCOOH] + [COOH-]) is to be 0.100M.

c) Calculate the pH of the buffer after 0.005 mol NaOH is added to 500mL of the buffer from part b.

d) Calculate the pH of 0.005 mol NaOH added to 500mL of water and compare the result to part c. How does the comparision illustrate the function of a buffer?

e) What is the pH of the buffer after 0.005 mol HNO3 is added to 500mL of the buffer from part b?

f) Calculate the pH of 0.005 mol HNO3 added to 500mL of water and compare the result to part e. How does the comparison illustrate the function of a buffer?

Explanation / Answer

a)

HCOOH ---> H+ + HCOO-

NaCOOH ---> Na+ + HCOO-

b) pH = 3.9 = pka + log [Salt / acid]

Pka = 3.74

so 0.16 = log [Salt / Acid]

1.44 = [salt / acid]

and salt + acid = 0.1

so 1.44 = salt / 0.1-salt

0.144 - 1.44 salt = salt

[Salt] = 0.059

[acid] = 0.041

c) molarity of NaOH = 0.005 X 1000 / 500 = 0.01

it will increase the conc of salt and decrease conc of acid

pH = 3.74 + log [0.059 + .01 / 0.041 - 0.01] = 4.08

d) The pH of NaOH ?

OH- = 0.01

so pOH = 2

so pH = 12

e) if acid is added the conc of acid = 0.01

it will increase the conc of acid and decrease the conc of salt

pH = 3.74 + log [0.049 / 0.051] = 3.72

f)

pH of HNO3 = -log[H+]

so pH= 2

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