Question

The pH of 0.50 M benzoic acid is 2.24. Calculate the change in pH when 2.64 g of C6H5COONa is added to 38 mL of 0.50 M benzoic acid, C6H5COOH. Ignore any changes in volume. The Ka value for C6H5COOH is 6.5 x 10-5.

Answer 1

Answer:

1.93

Explanation:

Moles of C$_{6}$H$_{5}$COOH = 38/1000 × 0.50 = 0.019mol

Moles of C$_{6}$H$_{5}$COONa = Mass/Molar mass = 2.64/144.10 = 0.018321mol

Final pH = pKa + log([C$_{6}$H$_{5}$COONa]/[C$_{6}$H$_{5}$COOH]

= -log Ka  + log(mols of C$_{6}$H$_{5}$COONa]/mols of C$_{6}$H$_{5}$COOH

= -log(6.5 × 10^(-5)) + log (0.018321/0.019)=4.17

change in pH = final - initial pH

= 4.17 - 2.24

=1.93