Answer to Question #192816 in Biochemistry for Jasmin

Question

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

Answer 1

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$\to$

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