Answer:
9.3 mL of 6 M acetic acid needs to be added to 500 mL of a solution of 0.20 M sodium acetate to a achieve a pH of 5.0
Explanation:
This problem can be solved by the Henderson-Hasselbalch equation. It's formula is:
[tex]pH=pKa+log(\frac{[CH_3COO^-]}{[CH_3COOH]})[/tex]
The molar concentration can be replaced by the moles of the solute as the volume of the buffer will be the same for both species
[tex]pH=pKa+log(\frac{n_{CH_3COO^-}}{n_{CH_3COOH}})[/tex]
Placing the given data:
[tex]5.0=4.75+log(\frac{0.1}{n_{CH_3COOH}})[/tex]
[tex]n_{CH_3COOH}=(\frac{0.1}{10^{5.0-4.75}})\\\\n_{CH_3COOH}=(\frac{0.10}{1.778})\\\\ n_{CH_3COOH}=0.056moles[/tex]
The volume required to obtain the above-calculated moles can be determined by the molarity of acetic acid
[tex]M_{CH_3COOH}=\frac{n_{CH_3COOH}}{V_{CH_3COOH}(L)}\\\\V_{CH_3COOH}=\frac{n_{CH_3COOH}}{M_{CH_3COOH}}\\\\V_{CH_3COOH}=\frac{0.056}{6.0}\\\\V_{CH_3COOH}=0.0093L\\\\or\\\\V_{CH_3COOH}=9.3mL[/tex]
The above volume of acetic acid can be added to 500 mL of acetate to form 5.0 pH buffer
