Answer to Question #192816 in Biochemistry for Jasmin

Question #192816

You have prepared a 0.11 M acetate buffer solution in a volume of 1 L at pH 5.76. You then added 80 mL of 0.1 M HCl to the solution. What is the final pH?

pKa of acetic acid: 4.76


1
Expert's answer
2021-05-14T06:33:02-0400

given, 0.11M acetate buffer solution of 1L. Hence, in 1L the amount of acetate=0.11 moles

Now,

80ml of 0.1M Hcl solution is added

so, in 80 ml or "80\\times10^{-3}L" Hcl solution contains

"\\to0.1\\times80\\times10^{-3}\\space moles\\\\\\to8\\times10^{-3}\\space moles"

Now,

"CH_3COO^-+H^+\\rightleftharpoons CH_3COOH"

so, "8\\times10^{-3}" moles will react with "8\\times 10^{-3}" moles of "CH_3C)OO^-" to produce same amount of "CH_3COOH\\to"

moles of "CH_3COOH\\to8\\times10^{-3}\\space moles"

concentration of "CH_3COOH\\to8\\times10^{-3}\\space moles"

concentration of "CH_3COOH\\to7.407\\times10^{-3}M"

"[CH_3COOH]=\\frac{8\\times10^{-3}\\times1}{1.08}=7.407\\times10^{-3},\\space total\\space vol=1.08L]"

Now,

"CH_3COO^-\\space remaining\\space moles\\\\=0.11-8\\times10^{-3}=0.102\\space ,moles"

concentration of "CH_3COO^-\\to\\frac{0.102\\times1}{1.08}=0.0944M"


According to Henderson"\\to"

"\\\\PH=Pka+log\\frac{[CH_3COO^-]}{[CH_3COOH]}\\\\PH=4.76+log\\frac{0.0944}{7.407\\times10^{-3}}=5.86\\\\so,\\space [Final \\space PH=5.86]"



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