How many grams of sodium benzoate, C H CO Na, have to be added to 1.50 L of a 0.0200 M solution of benzoic acid, C H CO H, to make a buffer with a pH = 4.00, assuming no volume change?

2.8275grams

Explanation: A buffer is made btw a weak acid and it salt. In a solution made by dissolving a weak acid in solution, equilibrium is set up btw ionised and unionised ion. For Benzoic acid

C6H5COOH> C6H5COO- + H+

Ka = [C6H5COO-] [H+]/ [C6H5COOH](1)

using Ka = 6.5× 10^-5, [C6H5COOH] = 0.02M. PH= - log[H+] > [H+]= 10^-4M.

Putting the values in(1)

[C6H5COO-]= 6.5× 10^-5 × 0.02/ 10^-4

[C6H5COO-] = 0.013M = Molarity of sodium benzoate

Mole(C6H5COONa) = 0.013 × Volume = 0.013mol/litre × 1.5 litre

Mole(C6H5COONa) = 0.0195mol

Mass(C6H5COONa) = 0.0195 × Molar mass

Mass(C6H5COONa) = 2.8275g

2b2t

explanation:

2b2t