Find the pH of a 0.25 M solution of lactic acid, HC3H5O3 (Ka = 8.3 x 10 -4 ).

pH → 1.83

Explanation:

Let's write the lactic acid as HLac, because it is a monoprotic weak acid.

HLac + H₂O  ⇄  H₃O⁺  +  Lac⁻      Ka

Initially we have 0.25 moles, so x amount has reacted. At the end, when the equilibrium is finished, we may have (0.25 - x) moles of acid, x moles of protons have been released and x moles of lactate were formed.

In order to find x, we use the acid constant, Ka. The expression for Ka is:

Ka = [H₃O⁺] .  [Lac⁻] / [HLac]

8.3×10⁻⁴  = x² / (0.25-x)

8.3×10⁻⁴ (0.25 - x) = x²

8.3×10⁻⁴ . 0.25 - 8.3×10⁻⁴x - x² → this is a quadractic function

a = -1,  b = 8.3×10⁻⁴, c = 2.07×10⁻⁴

We solve: (-b - √(b² - 4ac) / (2a)

(-8.3×10⁻⁴ - (√ ((8.3×10⁻⁴)² - 4 (-1) (2.07×10⁻⁴) ) / (2 . -1) = 0.0148

[H₃O⁺] = 0.0148 M

- log [H₃O⁺] = pH → - log 0.0148 = 1.83

increased stability in the completion of fluorine's outer shell.

it is the most commonly accepted one