calculate the ph after the addition of 100.0 mL of 1.0M HCL to a 500 ml...

Question

Question

calculate the ph after the addition of 100.0 mL of 1.0M HCL to a 500 ml buffer made up of 0.500M HF (ka=6.8x10^-4) & 0.500 M NaF?

Answer 1

Answer #1

initially

millimoles of HF = 0.500 x 500 = 250

millimoles of NaF = 0.500 x 500 = 250

millimoles of HCl added = 1.0 x 100 = 100

after HCl added

millimoles of HF = 250 + 100 = 350

millimoles of NaF = 250 - 100 = 150

total volume = 100 + 500 = 600 mL

[HF] = 350 / 600 = 0.583 M

[NaF] = 150 / 600 = 0.25 M

mixture of HF and NaF act as acidic buffer

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

pKa = - log Ka = - log 6.8 x 10^-4 = 3.17

pH = 3.17 + log [0.25] / [0.583]

pH = 2.80

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Answer 2

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