Question

i know the answer but i dont know how to get it

10. How many milliliters of 10.0 M HNO3 solution must be added to a LOL buffer system that is 0.001 M acetic acid (HC2H302) and 0.1 M sodium acetate (NaC2H302) to bring the pH to 5.452 The pka of acetic acid is 4.75 a. 1.58 ml d. 2.99 mL b. 3.92 mL e. It is impossible to tell from the given information c. 25.00 mL

Answer 1

10.Ans :-

Let moles of HNO3 added = y L

Moles of HNO3 = Molarity x Volume in L

= 10.0 M x y L

= 10 y mol

and

Moles of acetic acid acid = 0.001 M x 1.0 L = 0.001 mol

Moles of sodium acetate = 0.1 M x 1.0 L = 0.1mol

ICF table is :

...........................CH3COO- ..................+.................H+ ----------------> CH3COOH

Initial.....................0.1 mol ..................................10 y mol ................0.001 mol.............................

Change................-10 y mol..............................-10ymol..................+10 y mol........................

Final.....................(0.1-10y) mol .....................0.0 mol....................(0.001+10y)...................

Using Henderson-Hasselbalch equation :

pH = pKa + log [CH3COO-]/[CH3COOH]

5.45 = 4.74 + log (0.1-10y) / (0.001+10y)

log (0.1-10y) / (0.001+10y) = 0.71

(0.1-10y) / (0.001+10y) = 100.71

(0.1-10y) / (0.001+10y) = 5.13

(0.1-10y) = 5.13 (0.001+10y)

0.1 - 10y = 0.00513 + 51.3y

51.3y + 10y = 0.1 - 0.00513

61.3y = 0.09487

y = 0.09487/61.3

y = 0.0015476 L = 1.58 mL (approx.)

Therefore, volume of HNO3 required in mL = 1.58 mL

So, Option (a) is the answer.