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 4.60 ?

Answer 1

5.4 ml of 10 M HNO₃ added

: : By using the mode Henderson Solve this Hasselbalch extremium problem PH = pka + log [A-B/LHAJ : : Number of : : moles of Acid [HA] = Concentration x volume = 0.010 mol/l XI.OL = a.al mol = 0.6 mol moles of acetate [A] = ool x = Ool mole : : Number of : males of : Number of HNO3 added is fixed as a = x do Henderson expression become : . : + PH=46 pka - 425 (for Aceric oko . .. [A] - x PH = pka + 4 CHA] the 46= 4075 + log (2001)-x 001-X 406 = 4075 . 1-X ) Acid) Oo) - -0.15 = log on 10 10-015 = Oilure = 07079 001- x = 0.0544 mol volume of HNO3 = 1/c = 0.0544/10M = 0.00544 = 5.44 X10"L = 5:44 ml ans

EXPLANATION