Question

In: Chemistry

A vitamin C tablet containing 250 mg of ascorbic acid (C6H8O6; Ka=8.0

Expert Solution

We have , 250 mg of ascorabicacid (C6H8O6) dissolved in 250ml water . Ka = 8.0*10^-5

We know that , M = wt / mwt * 1000/v

= 0.250/176.12 * 1000/250

Intial concentration of ascorbic acid is , = 5.68 * 10^-3 moles/lit

Ascorbic acid dissolved in water , C6H8O6 +H2O ------------.> C6H7O6^- + H3O⁺

intial concentration --- 5.68*10^-3 0 0

concentration change ---- -x +x +x

equilibrium concentration ---- 5.68*10^-3 - x x x

Ka = [C6H7O6^-][H3O⁺] / [C6H8O6]

SO , 8 * 10^-5 = (X)(X) / ( 5.68*10^-3 - x )

8 * 10^-5 ( 5.68*10^-3 - x ) - X² = 0

X² + 8 * 10^-5 X - 4.54*10^-7 =0

X = -b b²-4ac / 2a

X = -(8*10^-5) (8*10^-5)² - 4(1)(- 4.54*10^-7) / 2(1)

[H3O⁺] = X = 6.3 * 10^-4 mol/lit

Since , [H3O⁺] = [C6H7O6^-] = 6.3 * 10^-4 mol/lit

pH = -log[H3O⁺] = -log(6.3*10^-4)

pH = 3.2

queen_honey_blossom answered 5 months ago

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