If you need to prepare 125 mlof 0.15m sodium acetate buffer solution that has a ph of 5.60, how many grams of sodium acetate (mw = 82.03/g/mol)and how many grams of acetic acid(mw = 60.05/g/mol)are needed?

mass of acetic acid needed = =0.156 g

MAss of sodium acetate needed =1.54g

Explanation:

The molarity of sodium acetate is 0.15 M

The pH of buffer solution is calculated from Hendersen Hassalbalch's equation as:

Given

pH = 5.60

pKa of acetic acid = 4.74

Putting values:

[salt]=7.24[acid]

Moles of salt present = molarity X volume = 0.15X0.125=0.01875

moles of acid = 0.01875/7.24=0.0026

mass of acetic acid needed = moles X molar mass =0.0026X60.05=0.156 g

MAss of sodium acetate needed = moles X molar mass = 0.01875X82.03

                                                          =1.54g

the answer to this question is 3.