Answer :
Answer: The mass of sodium acetate that must be added is 30.23 grams
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Molarity of acetic acid solution = 0.200 M
Volume of solution = 1 L
Putting values in above equation, we get:
[tex]0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol[/tex]
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pH=pK_a+\log(\frac{[\text{salt}]}{[\text{acid}]})[/tex]
[tex]pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})[/tex]
We are given:
[tex]pK_a[/tex] = negative logarithm of acid dissociation constant of acetic acid = 4.74
[tex][CH_3COONa]=?mol[/tex]
[tex][CH_3COOH]=0.200mol[/tex]
pH = 5.00
Putting values in above equation, we get:
[tex]5=4.74+\log(\frac{[CH_3COONa]}{0.200})[/tex]
[tex][CH_3COONa]=0.364mol[/tex]
To calculate the mass of sodium acetate for given number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Molar mass of sodium acetate = 83.06 g/mol
Moles of sodium acetate = 0.364 moles
Putting values in above equation, we get:
[tex]0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g[/tex]
Hence, the mass of sodium acetate that must be added is 30.23 grams