Answer to Question #192815 in General Chemistry for Jasmine

Question

How much acetic acid (of a 17.4 M stock) and sodium acetate (in g) are needed to prepare 1 L of a 0.1 M acetate buffer at pH 5.3?

pKa of acetic acid: 4.76

Molecular mass of acetic acid: 60.05 g/mol

Molecular mass of sodium acetate: 82.03 g/mol

Answer 1

According to the question -

Given values are -

the mass of sodium acetate neeeded to prepare 1 L of 0.1 M sodium acetate.

$Molarity=\frac{Moles of solute}{Volume in litres }$

                

 The molarity of the given solution is 0.1 M that means 0.1 mole of sodium acetate is dissolved in 1 liter. The molar mass of sodium acetate is 82 g/mol therefore 0.1 mole will be

      mass of 1 mole of sodium acetate =82g

$=\frac{0.1mol×82g}{1mol}$

=8.2g

The Henderson-Hasselbalch for pH of the buffer solution is

pH=pKa+Log$\frac{salt}{acid}$

by plugging in the data in the abve expression for pH

5.3=4.76+log$\frac{0.1}{acid}$

0.54=log$\frac{0.1}{acid}$

0.54   = log 0.1 M - log [acid]

    0.54  = -1.000 M - log [acid]

Add log[acid] on both sides

0.54+log[acid]=-1.000M

log [acid] = -1.540 M

    [acid]     = 0.0288 M

Concentration of acid needed    = 0.0288 M

volume of acetic acid needed.

This can be found by using dilution law

                 C₁V₁    = C₂ V₂

V₁=$\frac{0.0288M×1000mL}{17.4M}=1.66mL$

Mass of sodium acetate needed  = 8.2 g and

  Volume of stock solution of acetic acid needed = 1.66 mL