# Question cbc00

Jan 31, 2015

The final concentration of citrate in the buffer will be $\text{48.0 mmol/L}$.

Citric acid is a weak acid that dissociates into citrate (${C}_{6} {H}_{5} {O}_{7}^{3 -}$) in three steps (this takes place in aqueous solution)

${C}_{6} {H}_{8} {O}_{7} r i g h t \le f t h a r p \infty n s {H}^{+} + {C}_{6} {H}_{7} {O}_{7}^{-}$, ${K}_{a 1} = 7.4 \cdot {10}^{- 4}$

${C}_{6} {H}_{5} {O}_{7}^{-} r i g h t \le f t h a r p \infty n s {H}^{+} + {C}_{6} {H}_{6} {O}_{6}^{2 -}$, ${K}_{a 2} = 1.7 \cdot {10}^{- 5}$

${C}_{6} {H}_{6} {O}_{6}^{2 -} r i g h t \le f t h a r p \infty n s {H}^{+} + {C}_{6} {H}_{5} {O}_{7}^{3 -}$, ${K}_{a 3} = 4.0 \cdot {10}^{- 7}$

As you can see, the acid dissociation constants are smaller with every step, which means that the concentration of citrate will be negligible in solution.

SIDE NOTE: you should try and calculate this concentration, it should come out to be $\text{0.000021 mmol/L}$; the starting concentration for citric acid will be $\text{10.0 mmol/L}$.

This means that the added citric acid will not contribute to the concentration of citrate. What will contribute however will be sodium citrate (${C}_{6} {H}_{5} {O}_{7} N {a}_{3}$).

When placed in aqueous solution, sodium citrate dissociates into

${C}_{6} {H}_{5} {O}_{7} N {a}_{3} \to 3 N {a}^{+} + {C}_{6} {H}_{5} {O}_{7}^{3 -}$

One mole of sodium citrate will produce 3 moles of sodium cations and 1 mole of citrate in solution. So, the number of moles of sodium citrate you have is

n_("sodium citrate") = C * V = "60.0 mmol/L" * 200 * 10^(-3)"L"

n_("sodium citrate") = "12.0 mmol"

The final concentration of sodium citrate will be

${C}_{\text{sodium citrate") = n_("sodium citrate")/V_("total") = ("12.0 mmol")/(("200 + 50") * 10^(-3)"L}}$

C_("sodium citrate") = "48.0 mmol/L"

This means that the concentration of citrate will be

C_("citrate") = "48.0 mmol/L"

If you add the concentration of citrate that results from the three-step dissociation of citric acid, the final concentration will be

C_("final") = "48.0 mM" + "0.000021 mM" = "48.000021 mM"#

For all intended purposes, this concentration is equal to $\text{48.0 mM}$.