Homework Answers
Hydrozoic acid = N3H = 183.0 mL of 0.30M
number of mole of N3H = 0.30M x 0.1830L = 0.0549 moles
Ka = 1.9 x10^-5
-log(Ka) = -log(1.9 x10^-5)
Pka = 4.72
NaOH = 78.0 mL of 0.18M
numberr of moles of NaOH = 0.18M x 0.0780L = 0.01404 moles
N3H + NaOH --------------------- N3Na + H2O
Initial 0.0549 0.01404 0
change - 0.01404 - 0.01404 + 0.01404
remaining 0.04086 0 + 0.01404
PH = Pka + log[salt]/[acid]
PH = 4.72 + log( 0.01404/0.04086)
PH = 4.26
The answer is A.
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