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A buffer is prepared by adding 42.1 mL of 2.4 M NaF to 30.3 mL of...

A buffer is prepared by adding 42.1 mL of 2.4 M NaF to 30.3 mL of 0.18 M HF. What is the pH of the final solution?

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Answer #1

We first calculate the molarity of each component in the final 72.4mL = 0.0724L solution:
Mol NaF in 42.1mL of 2.4M solution = 42.1/1000*2.4 = 0.0175 mol
Molarity = 0.0175/0.0724 = 0.241M NaF
Mol HF in 30.3mL of 0.18M solution = 30.3/1000*0.18 = 0.168 mol
Molarity = 0.168/0.0724 = 2.32 M HF
pKa HF = - log Ka = - log 7.2 × 10-4 = 3.14
Use the H-H equation:
pH = pKa + log ( [salt]/[acid])
pH = 3.14 + log (0.241/2.32)
pH = 3.14 + log 0.103
pH = 3.14 + (-0.98)
pH = 2.16.

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