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.

1.93

Explanation:

Moles of CHCOOH = 38/1000 × 0.50 = 0.019mol

Moles of CHCOONa = Mass/Molar mass = 2.64/144.10 = 0.018321mol

Final pH = pKa + log([CHCOONa]/[CHCOOH]

= -log Ka  + log(mols of CHCOONa]/mols of CHCOOH

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

change in pH = final - initial pH

= 4.17 - 2.24

=1.93

the answer is b. dilution

the answer is d. 74.02