You prepared a solution using 4.0mL of .10M CuSO 4 ( copper (ll) sulfate) and 5.

Question

You prepared a solution using 4.0mL of .10M CuSO4 (copper (ll) sulfate) and 5.0mL of .10M Na2C4H4O6 (sodium tartrate), lastly you added 1.0mL of deionized water for a total volume of 10.0mL. Using a linear equation from a Beer’s Law plot, you determined the equilibrium concentration of the Cu2+ ions in the saturated copper (II) tartrate solution to be 2.0x10-4moles of Cu2+ions. Now that the equilibrium concentration of the copper ions is known, set up an ICE table to determine the equilibrium concentration of the tartrate. The tartrate concentration at equilibrium is different than the copper ion concentration since there were initially more moles of tartrate in the mixture.

Equilibrium concentration of C4H4O62-in solution______________________?

I have to repeat this process a few times so please show how you calculated the answer. Thank you

Explanation / Answer

Initial moles of tartarate = molarity x volume = 4 x 10^-4 mols

intial moles of CuSO4 = 0.1 x 0.005 = 5 x 10^-4 mols

set up an ICE table,

                         Cu2+                  +        tartarate    --------->    Cu(taratrate)

initial               4 x 10^-4                         5 x 10^-4                              0

change           -2 x 10^-4                        -2 x 10^-4                      2 x 10^-4

equilibrium  (4x10^-4 - 2x10^-4)      (5x10^-4 - 2x10^-4)             2 x 10^-4                                                       

So the equilibrium concentration of tartarate becomes = 3 x 10^-4/0.01 = 0.03 M

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