103 398
Assignments Done
98.5%
Successfully Done
In November 2021

# 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
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]

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!