9. What is the pH of a solution of 0.200M H3C6H5O7 (citric acid)? Ka = 3.5 x 10-4

Ionization equilibrium: H3C6H5O7 (aq) H+ (aq) + H2C6H 5O7- (aq)

I.C.E. table:

H3C6H5O7 (aq) H+ (aq) H2C6H5O7- (aq)
INTIAL
0.200M
0M
0M
CHANGE
-xM
+xM
+xM
EQUILIBRIUM
(0.200-x)M
xM
xM

Ka=3.5 x 10-4 = {x2} / {(0.200-x)}

To solve the expression for x, what kind of an assumption do we need to make and why?
a) x >> 0.200 ; to get rid of the x2 term
b) x << 0.200 ; to avoid the quadratic equation
c) x >> 3.5 x 10-4 ; to eliminate the x2 term
d) x << 3.5 x 10-4 ; to avoid a quadratic equation