If 8.72 grams of oxalic acid, H2C2O4, a diprotic acid, is neutralized when 23.4 mL of a solution of KOH is added, what is the molarity of KOH?

the topic is acid-base triation, and i have already converted the mass to moles,but I am having trouble finding the volume for the acid (H2C2O4), I am guessing it's just 8.72 mL? Because grams=mL, right? to calculate the molarity i would just divide the moles by volume(in Liters)

I think that's no and yes but here is what you do. All of these are done stepwise and most follow the same pattern.

1. Write the equation.
H2C2O4 + 2KOH ==> K2C2O4 + 2H2O

2. Convert g H2C2O4 to mols. mols = grams/molar mass.

3. Using the coefficients in the balanced equation, convert mols H2C2O4 to mols KOH. That's the value in #2 x 2.

4. Now M KOH = mols KOH/L KOH

You might do well to memorize this four-step process. It will work many problems for you.

To find the volume of the acid (H2C2O4), we cannot assume that the volume is the same as the mass in grams. While mass and volume can be equal for certain substances like water (1 mL = 1 g), it is not true for all substances, especially when dealing with solid compounds like oxalic acid.

To find the volume of the oxalic acid solution, you need to convert the given mass (8.72 grams) to moles using the molar mass of H2C2O4. The molar mass of H2C2O4 can be calculated by adding the atomic masses of each element: 2(1.01 g/mol) for hydrogen, 2(12.01 g/mol) for carbon, and 4(16.00 g/mol) for oxygen.

So, the molar mass of H2C2O4 = (2 × 1.01 g/mol) + (2 × 12.01 g/mol) + (4 × 16.00 g/mol) = 90.03 g/mol.

Next, calculate the number of moles of oxalic acid by dividing the mass by the molar mass:

moles of H2C2O4 = mass / molar mass = 8.72 g / 90.03 g/mol.

Now that you have the moles of H2C2O4, you need to find the volume of the solution in liters. Knowing the volume of KOH added (23.4 mL), you can use the balanced equation for the neutralization reaction to calculate the ratio of moles of H2C2O4 to moles of KOH.

The balanced equation for the neutralization of oxalic acid by KOH is:

H2C2O4 + 2KOH → K2C2O4 + 2H2O.

From the equation, you can see that for every 2 moles of KOH, 1 mole of H2C2O4 is neutralized. This means the moles of KOH used is twice the moles of H2C2O4.

moles of KOH = 2 × moles of H2C2O4.

Now, divide the moles of KOH by the volume of KOH solution in liters to find the molarity (M) of KOH:

molarity of KOH = moles of KOH / volume of KOH solution (in L) = (2 × moles of H2C2O4) / 0.0234 L.

By plugging in the moles of H2C2O4 and the given volume of KOH solution, you will get the molarity of KOH.