<u>Answer:</u> The mass of sodium acetate that must be added is 30.23 grams
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
![\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20the%20solution%7D%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7BVolume%20of%20solution%20%28in%20L%29%7D%7D)
Molarity of acetic acid solution = 0.200 M
Volume of solution = 1 L
Putting values in above equation, we get:
![0.200M=\frac{\text{Moles of acetic acid}}{1L}\\\\\text{Moles of acetic acid}=(0.200mol/L\times 1L)=0.200mol](https://tex.z-dn.net/?f=0.200M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20acetic%20acid%7D%7D%7B1L%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20acetic%20acid%7D%3D%280.200mol%2FL%5Ctimes%201L%29%3D0.200mol)
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[CH_3COONa]}{[CH_3COOH]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5BCH_3COONa%5D%7D%7B%5BCH_3COOH%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of acetic acid = 4.74
![[CH_3COOH]=0.200mol](https://tex.z-dn.net/?f=%5BCH_3COOH%5D%3D0.200mol)
pH = 5.00
Putting values in above equation, we get:
![5=4.74+\log(\frac{[CH_3COONa]}{0.200})](https://tex.z-dn.net/?f=5%3D4.74%2B%5Clog%28%5Cfrac%7B%5BCH_3COONa%5D%7D%7B0.200%7D%29)
![[CH_3COONa]=0.364mol](https://tex.z-dn.net/?f=%5BCH_3COONa%5D%3D0.364mol)
To calculate the mass of sodium acetate for given number of moles, we use the equation:
![\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}](https://tex.z-dn.net/?f=%5Ctext%7BNumber%20of%20moles%7D%3D%5Cfrac%7B%5Ctext%7BGiven%20mass%7D%7D%7B%5Ctext%7BMolar%20mass%7D%7D)
Molar mass of sodium acetate = 83.06 g/mol
Moles of sodium acetate = 0.364 moles
Putting values in above equation, we get:
![0.364mol=\frac{\text{Mass of sodium acetate}}{83.06g/mol}\\\\\text{Mass of sodium acetate}=(0.364mol\times 83.06g/mol)=30.23g](https://tex.z-dn.net/?f=0.364mol%3D%5Cfrac%7B%5Ctext%7BMass%20of%20sodium%20acetate%7D%7D%7B83.06g%2Fmol%7D%5C%5C%5C%5C%5Ctext%7BMass%20of%20sodium%20acetate%7D%3D%280.364mol%5Ctimes%2083.06g%2Fmol%29%3D30.23g)
Hence, the mass of sodium acetate that must be added is 30.23 grams