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Ascorbic acid (Vitamin C) is a diprotic acid, H₂C6H6O6. What is the pH of a 0.10 M
solution? What will be the concentration of C6H6O62-? Ka1 = 7.9 x 105 and Ka2 = 1.6 x 10-12

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The pH of the solution is 1.52 and the concentration of C6H6O62- is 2.2 * 10^-7 M.

What is the concentration of  C6H6O62-?

We know that;

H₂C6H6O6 ⇔HC6H6O6^-(aq)    +   H^+(aq) ----- (1)

Ka1 = 7.9 x 10^-5

Again;

HC6H6O6^-(aq) ⇔ C6H6O6^2-(aq)  +   H^+(aq) ------- (2)

Ka2 = 1.6 x 10^-12

Given that Ka2 <<<< Ka1 we have to apply equation 1 to obtain H^+ hence;

Ka1 = [HC6H6O6^-] [H^+]/[H₂C6H6O6]

Let  [HC6H6O6^-] = [H^+] =x

7.9 x 10^5 =x^2/0.1 - x

But x <<< 0.1 thus;

7.9 x 10^5 =x^2/0.1

x = √7.9 x 10^-5  * 0.1

x = 0.03 M

pH = -log(0.03 M)

pH = 1.52

To obtain  C6H6O6^2-;

Ka2 = [ C6H6O6^2-] [ H^+]/[HC6H6O6^-]

Also;

[ C6H6O6^2-] = [ H^+]= x

Now;

1.6 x 10^-12 = x^2/0.03 -x

But x<<<0.3

Hence;

1.6 x 10^-12 = x^2/0.03

x = √1.6 x 10^-12 * 0.03

x = 2.2 * 10^-7 M

The concentration of C6H6O6^2- is  2.2 * 10^-7 M.

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