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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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HamidTaher HamidTaher
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3 years ago
When 2.4 × 10–2 mol of nicotinic acid (a monoprotic acid) is dissolved in 350 mL of water, the pH is 3.05. Calculate the Ka of nicotinic acid.


▸ 1.3 × 10–2

▸ 1.2 × 10–5

▸ 6.8 × 10–2

▸ 3.4 × 10–5

▸ None of these are correct.
Textbook 
Chemistry: An Atoms First Approach

Chemistry: An Atoms First Approach


Edition: 3rd
Authors:
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Josh D.Josh D.
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#2
A year ago
When 1.5 × 10–2 mol of nicotinic acid (a monoprotic acid) is dissolved in 350 mL of water, the pH is 3.05. Calculate the Ka of nicotinic acid.
wrote...
#3
Educator
A year ago
Quote from: gaby.garcia5141 (A year ago)
When 1.5 × 10–2 mol of nicotinic acid (a monoprotic acid) is dissolved in 350 mL of water, the pH is 3.05. Calculate the Ka of nicotinic acid.

concentration of nicotinic acid=[nicotinic acid]=1.5*10^-2 mol/0.350 L=4.3*10^-2 mol/L

pH=-log[H3O+]=3.05

[H3O+]=10^-3.05=8.9*10^-4 M

nicotinic acid +H2O<--->conjugate base of nicotinic acid+H3O+

ka=[base][H3O+]/[nicotinic acid]

[base]=[H3O+]=8.9*10^-4 M

ka=(8.9*10^-4 M)^2/(4.3*10^-2 M)=18.42*10^-6=1.84*10^-5

ka=1.84*10^-5
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