A sample of KHC2O4 weighing 0.5839 g is dissolved in 25 mL of water and an endpoint is reached when 31.99 mL of KOH has been added. Calculate the molarity of the KOH solution.

KHC2O4(aq) + KOH(aq) �¨ K2C2O4(aq) + H2O(l)

because the coefficents are all the same, i just assumed that KOH also had 0.5839 g, converted into mols, and divdied by the total volume. however, this answer was wrong...

I'm not surprised the answer was wrong.

Chemical reactions are by mols and not by grams; therefore, 1 mol of each reactant produces 1 mol of each product but a mol is not a gram.
SO, convert 0.5839 g KHC2O4 to mols, convert mols KHC2O4 to mols KOH (that's the 1:1 part), then calculate the molarity of KOH. Post your work if you get stuck. I expect you will have no trouble.

To calculate the molarity of the KOH solution, you need to use the balanced equation for the reaction and the given information.

First, calculate the number of moles of KHC2O4. We have the mass of KHC2O4, which is 0.5839 g. The molar mass of KHC2O4 is 128.13 g/mol. Therefore, the number of moles of KHC2O4 can be calculated as follows:

moles of KHC2O4 = mass of KHC2O4 / molar mass of KHC2O4
= 0.5839 g / 128.13 g/mol
= 0.004551 mol

Next, we need to determine the number of moles of KOH based on the volume of KOH added. Given that the volume of KOH solution added is 31.99 mL, we need to convert it to liters. There are 1000 mL in 1 L, so:

volume of KOH solution = 31.99 mL * (1 L / 1000 mL)
= 0.03199 L

Now, we can calculate the molarity of the KOH solution using the moles of KOH and the volume of the solution:

molarity of KOH = moles of KOH / volume of KOH solution
= 0.004551 mol / 0.03199 L
= 0.142 mol/L

Therefore, the molarity of the KOH solution is 0.142 mol/L.