Question

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?

Answer 1

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)`