When 0.100 L of 0.200 M K2C2O4 react with 0.150 L of 0.250 M BaBr2, BaC2O4 forms an insoluble precipitate. Calculate the final concentrations of the ions that remain in solution after the reaction.
I know how to get the net ionic equation (Ba2+ + C2O4- -> BaC2O4).
HOWEVER - I don't know why the three ions that remain in solution are K+, Br-, and Ba2+. Where did K+ come from? Thanks!
K, Br are ions because the compounds formed go into solution as ions.
Ba ion is there because there is excess Ba ions. Look at the balanced equation.
To understand why K+ is one of the ions that remain in solution after the reaction, we need to examine the reactants involved.
In the given reaction, K2C2O4 (potassium oxalate) reacts with BaBr2 (barium bromide) to form BaC2O4 (barium oxalate), which is insoluble. This means that BaC2O4 will precipitate out of the solution as a solid.
Initially, you have 0.100 L of a 0.200 M solution of K2C2O4. This means that you have (0.200 mol/L) x (0.100 L) = 0.0200 mol of K2C2O4.
Similarly, you have 0.150 L of a 0.250 M solution of BaBr2. This means that you have (0.250 mol/L) x (0.150 L) = 0.0375 mol of BaBr2.
Now, let's look at the reaction equation:
K2C2O4(aq) + BaBr2(aq) -> BaC2O4(s) + 2KBr(aq)
From the balanced equation, we can see that K2C2O4 reacts with BaBr2 to form BaC2O4, which precipitates out as a solid. At the same time, two moles of KBr are formed for every mole of K2C2O4.
Since we initially have 0.0200 mol of K2C2O4, it will react completely to form 2 x 0.0200 mol = 0.0400 mol of KBr. This means that after the reaction, you will have 0.0400 mol of K+ ions in solution.
As for the Ba2+ and Br- ions, we initially have 0.0375 mol of BaBr2. Since the ratio in the balanced equation is 1:1, this means we will have 0.0375 mol of Ba2+ ions and 0.0375 mol of Br- ions after the reaction.
To calculate the final concentrations, you need to divide the number of moles by the final volume of the solution. In this case, the final volume is given as the sum of the initial volumes of K2C2O4 and BaBr2, which is 0.100 L + 0.150 L = 0.250 L.
Consequently, the final concentration of K+ ions in solution after the reaction is 0.0400 mol / 0.250 L = 0.160 M. The final concentration of Ba2+ and Br- ions is 0.0375 mol / 0.250 L = 0.150 M for each ion.
Therefore, the final concentrations of the ions that remain in solution after the reaction are K+ (0.160 M), Ba2+ (0.150 M), and Br- (0.150 M).