Question

a) What is the pH of a buffer made by combining 0.025 moles potassium acetate with 0.015 moles acetic acid in 200 mL of water? 8.97 b)What is the new pH if 0.5 mL of 10 M HCl is added to the buffer in part a? 9.20

Answer 1

(a) CH₃COOH Kay CH₃Coo +ht ka= 1.8x105 PH of this buffer mixture will be calculated by using Handerson Hasselbalch Equation : PH = Pka + log [CH₂600] [CH₃COOH] Concentration of Potassium acetate - Moles Volume (L) Molus = 0.025 meles volume = 200mL = 0.2L. [CH₂600] = 0.02 mol = 0-125M. 0.2L Concentration of acetic acid (CH3COOH) = 0.015 mel 0.2L = 0.075M Pka = -log ka = -log (1-8x105) = 4.74 PH = 4.74 + log 0.125m 0.075 M. - 4.74 + log (1666) = 4.74 +0.22 IPH = 4.96) After the addition of Hee concentration of CH₂COOH increases and that of CH₂C00 will decrease by aeron CH₃Coo & HU CH₃COOH + ce- (6) equation :

Moles of Hel = concentration xvolume (2) = 10m x 0.5 mL x 1032 ImL. = 0.005 mel. Сн, Coo + H+ + снасоон. I 0.025 0.005 0.015 mel C -0.005 -0.005 +0.005 E 0.020 0 0.010 Volume change is negligible after addition of 0.6 mL of Hal. PH = Pka + log [CH3 C00] [CH₂ coor] = 4.74 + 109 (0.020 x:20 M) 16.010 x0,20m) = 4.74 + log 2. PH = 4.74 +0.3. [PH = 5.04