Question

I need help on this please I just need it to be checked.

When 50 mL of a 0.10 M NaF aqueous solution is added to 50 mL of a 0.10 M HF aqueous solution, relative to the pH of the 0.10 M HF solution, the pH of the resulting solution will remain the same o become 7 increase decrease

Answer 1

We can solve given problem in following steps.

Step 1 : Calculation of pH of 0.1 M HF solution.

For weak acid we have relation, pH = 1/2 ( pKa - log C )

$\therefore$ pH of 0.1 M HF solution = 1/2 ( 3.17 - log 0.1 )

= 1/2 ( 3.17 -(-1))

= 1/2 ( 4.17 )

= 2.085

= 2.09

Step 2 : Calculation of pH of solution prepared by mixing 50 ml of 0.10 M NaF and 50 ml of 0.10 M HF.

Solution contain weak acid HF and its salt NaF. This solution acts as a buffer solution.

As the volume of solution doubles , concentrations will be half.

Therefore, [ NaF ]= 0.050 M and [ HF ] = 0.050 M

We have Henderson's equation, pH = pKa + log [ salt ] / [ acid ]

pH = pKa + log [ NaF ] / [ HF ]

pH = 3.17 + log 0.050 / 0.050

pH = 3.17 + 0

pH = 3.17

Step 3 : Comparison of pH of two solutions.

pH of 0.10 M HF solution is 2.09 and pH of soltuon prepared by mixing HF and NaF is 3.17.

pH of solution prepared by mixing HF and NaF is more than pH of 0.10 M HF solution.

ANSWER : Increase