- #1

i_love_science

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- Homework Statement
- Citric acid (H3C6H5O7) is a triprotic acid with Ka1 = 8.4*10^-4, Ka2 = 1.8*10^-5, and Ka3 = 4.0*10^-6. Calculate the pH of 0.15 M citric acid.

- Relevant Equations
- polyprotic acid equations (below)

Using the first equilibrium, H

Using the second equilibrium, H

The question asks to find the pH though, not specifically the concentration [H

Could anyone explain why, or whether my reasoning is right? Thanks.

_{3}A <-> H_{2}A^{-}+ H^{+}, the [H^{+}] is solved to be 0.01165 M ~ 1.1 * 10^-2 M.Using the second equilibrium, H

_{2}A^{-}<-> HA^{2-}+ H^{+}, the additional change in H^{+}is found to be 1.8 * 10^-5 M. The solution says that this second equilibrium is negligible, because 1.1 * 10^-2 M + 1.8 * 10^-5 M ~ 1.1 * 10^-2.The question asks to find the pH though, not specifically the concentration [H

^{+}]. I don't understand why it would not be better to add 0.01165 M + 1.8 * 10^-5 M = 0.011668 M, and continue with this type of calculation with the third equilibrium as well? Would it make the answer more precise? I got pH 1.93 using this method, but the solution gives 1.96.Could anyone explain why, or whether my reasoning is right? Thanks.