When 2.0 x 10-2 mole of nicotinic acid (amonoprotic acid) is dissolved in 350 mL of water, the pH is 3.05.What is the Ka of nicotinic acid?

Question

asked Dec 13, 2020 131k views

1 Answer

Demenvil · Answer 1 · 2020-12-18T13:12:00+0000

Answer:

Ka of nicotinic acid =
1.41 * 10^(-5)

Step-by-step explanation:

pH = 3.05

pH = -log [H^+]

H^+ = (10)^(-3.05)=0.00089125 M

No. of mol of nicotinic acid =
2.0 * 10^(-2)

Volume of water = 350 mL = 0.0350 L

Molarity =
$(Moles)/(Volume\ in\ L)$

Molarity =
$(2.0 * 10^(-2))/(0.350) = 0.05714\ M$

Nicotinic acid dissoctates as:

$HA \rightarrow H^+ + A^-$

[H+] = 0.00089125 M

[A-] = 0.00089125 M

[HA} at equilibium = 0.05714 - 0.00089125 = 0.05624875 M

Ka = ([H^+][A^-])/([HA])

Ka = ((0.00089125)^2)/(0.05624875) = 1.41 * 10^(-5)