Determine the pH of a solution that is 1.00 L of 0.100 M HF and 0.100 M NaF after 0.50 mole of solid KOH has been added to the solution. Ka(HF) = 3.5×10-4

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Question

Determine the pH of a solution that is 1.00 L of 0.100 M HF and 0.100 M NaF after 0.50 mole of solid KOH has been added to the solution. Ka(HF) = 3.5×10-4

Determine the pH of a solution that is 1.00 L of 0.100 M HF and 0.100 M NaF after 0.50 mole of solid KOH has been added to the solution. Ka(HF) = 3.5×10^-4

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Answer 1

Answer #1

mol of KOH added = 0.5 mol

HF will react with OH- to form F-

Before Reaction:

mol of F- = 0.1 M *1.0 L

mol of F- = 0.1 mol

mol of HF = 0.1 M *1.0 L

mol of HF = 0.1 mol

0.1 mol of HF and KOH would react.

So,

mol of KOH remaining = 0.4 mol

This is strong base which is still remaining

So,

[OH-] = mol of KOH / volume

= 0.4 mol / 1.00 L

= 0.4 M

use:

pOH = -log [OH-]

= -log (0.4)

= 0.3979

use:

PH = 14 - pOH

= 14 - 0.3979

= 13.6021

Answer: 13.60

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Answer 2

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