Question

i need help with number 4 and 5 please

4. How many mL of 0.100 M NaOH should be added to 500.0 mL of 0.150M acetic acid to produce a buffer pH 4.0? pKa acetic acid 4.757. ANSWER: 112 mL 5. How many mL of 0.100 M HCI should be added to 500.0 mL of 0.150 M sodium acetate to produce a buffer pH 5.0? pKa acetic acid 4.757 ANSWER: 273 mL

Answer 1

According to Henderson - Hasselbalch Equation : Aniyor Equation? CH₃COont Noon CH Coona & Ho (0.075-x) x 0.075 obtained from (6. 150 mol/L) x (0.500 L) = 0.075 mol by Equation : pH = pka + log [base/acid] 4.00 -4.757 + log %10.075-x3] - 0.157 = log [2/(0.015-x)] on 15-x) = Antilog (-0,157) = 6,175 x = 0,175 (0.075-x) x = 0.01313 - 0.1752 1.175x = 0.01313 x=0.01313 = 0,01117 noles Noon given 0,100M solution so for calculating volume in mL as, 0.01117 mol = 0.1117 L = 111.7mL 0.100 mol/L 1.75

Anys! hce & Cha Crona - Cug Coon of Nace x 6.015-x) 6.075 obtained from (0.150 mol/L) x (0.500L) = 0.075 mol pH = pka + log (base /acid] 5.0 24.057 + log (0.05-x] 0.243 = log (6075 ] 0,075-2 - Antilog (6.243) = 1.75 0,075-x = 1.75x 0.075 = 105x + x = 2.750 x = 0,075 = 0.0273 roles 2.75 ce given 0.100 Molar solution, so volume wed in me be 0.0273 mol = 0.272 L = 273 mL 0.100 nol/L