This sounds like homework, so I don't want to give too much away, but:

for the reaction H_{3}A <=> H_{2}A^{–} + H^{+}

10^{–pKa} = Ka = [H_{2}A^{–}]*[H^{+}]/[H_{3}A]

The values in square brackets are concentrations in molarity. Since one H^{+} and one H_{2}A^{–} is produced by the consumption of one H_{3}A, you can plug in the initial concentration of H_{3}A as whatever molarity you want (M), assume that there is no appreciable concentration of H^{+} or H_{2}A^{–}

so the calculation becomes

10^{–pKa} = x^{2}/[M–x]

solve for x, which is then the concentration of H^{+}, so pH = –log(x).

I have made the assumption that the citric acid is only deprotonated once. You can do the calculation taking all pKa values into account, but that is a LOT more work, and ends up giving a negligibly different answer.