Calculate the pH of a 4.988 M solution of ascorbic acid, which has a ka of 8.0 x 10-5. Report your answer with two decimal places. Example: 6.99

pH= 1.70

Explanation:

I      4.988        0     0

C     -x             +x     +x

E     4.988 -x    x       x

8.0 x 10^-5 = X^2/ 4.988 - x

(since the X value is so small, round 4.988-x to 4.988 because X is close to zero)

8.0 x 10^-5 = x^2/4.988

x4.988           x4.988

4.0 x 10^-4 = x^2

x = .0199

pH=-log[.0199] = 1.70

the ratio of the conjugate base to the acid required to attain the desired ph.