Question

How much 10.0 M HNO3 must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the pH to 5.05 ? The pKa of acetic acid is 4.75.

Answer 1

Hence, 2.67 ml of 10 M HNO3 must be added....

Answer 2

The pt of - Calulated equation, any butter solution can be using Henderson-Hasse bath pH = pkatlog [Acid] [ Base [ Base) - concentration of base. [Acid] = pka = -logna ka & Concentration of acid. Aud dissociation Constant Acetic acid. and CH₂ cooh is very weak Its dissociation can be writtenas, CH3COOH ca f t t can + CH₂ crocan . Acid Conjugare base) oka=4.75 ph before addition of JA NO 3. According to Henderson-tasselbalch equation, pH = pra thig I CH3C00-] | = P - 415 4 3 2 1 To.0100/ 7 pH = 4.75 +1 pH = prea tog [CH3 100h) 2 pH = 5.75 molarity nolarity = moles of Volume solute volume (4) . GEVCU VCL)

pH after Moles of addition of tmaz CH₂ cool in Tool solution - o.olmore IL x 12 soles of - CH₃coona 0.0 moie in 1.50 L solution = 0.1mole 11 214 = 0.1 male Suppose 'n' more of HNO3 must be added to reduce the pH. The dissociation equilibrium, ICE table Coltz coo-cay + HMI Can CH3COOH (y + Noz mitial oil 0.01 change, an th (0.olth) 9) Equilibrium (0.1-2) o According to ten derason Hasselbalch equation pH = pka las [CH3Coo- tlog [CH3 COOH + = 5.05 = 4.75 +109 0.1-n Ol- D'ont 0.0ltn pH = 5:05 0.01 + 100054.75 0.1-n Doo! tn 1.01952 0-1 - n = 0.01975 +1.9952 n 7,9952 ne 0:08 h = 0.02672 i holes of H Mag added 0.02 672 mole : volume of linaz added Jackman ( mole 0 Toa Reduired volume = 2.672 x 10°3L 2.672) mL concentratia 0.02 672 moles 10 mole 2.672x103/L - (2:672318!/LICE