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