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?

Ka of nicotinic acid =

Explanation:

pH = 3.05

No. of mol of nicotinic acid =

Volume of water = 350 mL = 0.0350 L

Molarity =

Molarity =

Nicotinic acid dissoctates as:

[H+] = 0.00089125 M

[A-] = 0.00089125 M

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

hi,

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