When a reaction mixture with a total volume of 1750 mL that is 0.0263 M aqueous K+ was stoichiometrically produced as per the balanced equation, what volume (mL) of 0.0326 M aqueous K3PO4 was required?

3 BaBr2(aq) + 2 K3PO4(aq) → Ba3(PO4)2(s) + 6 KBr(aq)

I would approach the problem this way.

0.0263 M x 0.1750 L = ?? moles K^+.

It will take ??moles K^+ x (2 mols K3PO4/6 moles K^+) = xx moles K3PO4.

Then M = moles/L. You know moles and M, solve for liters and multiply by 1000.
Actually, it is easier to work in millimoles.
0.263 x 1750/3 = millimoles K3PO4.
Then millimoles/mL = M and you solve for mL directly. Check my thinking (but don't invert the multiplier. ;-).

To find the volume of 0.0326 M aqueous K3PO4 required, we can use the stoichiometry of the balanced equation and the given concentration and volume of the aqueous K+ solution.

First, let's determine the number of moles of K+ in the 1750 mL of 0.0263 M K+ solution:
moles of K+ = concentration × volume
moles of K+ = 0.0263 mol/L × 1.75 L
moles of K+ = 0.0459 mol

Now, using the stoichiometry of the balanced equation, we can determine the amount of K3PO4 needed to react with the K+ ions. According to the balanced equation, the molar ratio of K+ to K3PO4 is 2:1.

moles of K3PO4 = 0.0459 mol K+ × (1 mol K3PO4 / 2 mol K+)
moles of K3PO4 = 0.0229 mol

Finally, we can calculate the volume of 0.0326 M K3PO4 solution needed to achieve this amount of moles.

volume of K3PO4 solution = moles of K3PO4 / concentration
volume of K3PO4 solution = 0.0229 mol / 0.0326 mol/L
volume of K3PO4 solution = 0.702 L

Convert the volume from liters to milliliters:

volume of K3PO4 solution = 0.702 L × 1000 mL/L
volume of K3PO4 solution = 702 mL

Therefore, approximately 702 mL of the 0.0326 M K3PO4 solution is required to produce a stoichiometric reaction mixture.