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.
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