12. How many grams of NaF must be added to 2.00 L of 0.100 M HF...

Question

Question

science chemistry

Answer 1

Answer #1

Ka = 6.8*10^-4

pKa = -log(6.8*10^-4) = 3.1675

pH = pKa + log [F-]/[HF]

pH - pKa = log [F-] / [HF]

4.00 - 3.1675 = log [F-]/ [HF]

0.8325 = log [F-]/ [HF]

10^0.8325 = [F-]/0.100

6.799 = [F-]/0.100

[F-] = 0.6799 M

moles NaF = 0.6799 M x 2.00 L= 1.3598

Mass = moles * molar mass

mass = 1.3598 * 41.98 = 57.09 g

option (e) is correct. (nearest value)

Answer 2

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