Question #cbc00

The final concentration of citrate in the buffer will be #"48.0 mmol/L"#.

Citric acid is a weak acid that dissociates into citrate (#C_6H_5O_7^(3-)#) in three steps (this takes place in aqueous solution)

#C_6H_8O_7 rightleftharpoons H^(+) + C_6H_7O_7^(-)#, #K_(a1) = 7.4 * 10^(-4)#

#C_6H_5O_7^(-) rightleftharpoons H^(+) + C_6H_6O_6^(2-)#, #K_(a2) = 1.7 * 10^(-5)#

#C_6H_6O_6^(2-) rightleftharpoons H^(+) + C_6H_5O_7^(3-)#, #K_(a3) = 4.0 * 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 #"0.000021 mmol/L"#; the starting concentration for citric acid will be #"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_6H_5O_7Na_3#).

When placed in aqueous solution, sodium citrate dissociates into

#C_6H_5O_7Na_3 -> 3Na^(+) + C_6H_5O_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_("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 #"48.0 mM"#.