Homework Answers
1)
pKa=- log Ka
pKa= - (log 4.2* 0^-4) = 4-.62 = 3.38
2)
pKa= - log Ka
pKa= - (log 3.27 * 10^-4)
= 4- 0.51 = 3.49
3)
pKa= 4.19
Concentration = 1 mg in 1 ml
In 1000 ml, 1 gram.
Number of moles in 1 liter (concentration)= 1/230 = 4.3 * 10^-3
Using Henderson Hasselbach equation
pH= pKa+ log (base/Acid)
pH= 4.19+ log (1/4.3*10^-3)
pH= 4.19+ 2.37
pH= 6.56
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