If it takes 20.0 mL of 1.200 M KOH to neutralize 60.0 mL of a given H3PO4 solution, what is the molarity of this phosphoric acid solution?
I've figured out that the neutralization of phosphoric acid by the addition of potassium hydroxide in aqueous solution: H3PO4 + 3 KOH -> K3PO4 +3 H2O
You've done step 1.
Step 2.
mols KOH = M x L KOH = ?
Step 3.
Using the coefficients in the balanced equation, convert mols KOH to mols H3PO4. You can see that mols H3PO4 = 1/3 mols KOH
Step 4.
Then M H3PO4 = mols H3PO4/L H3PO4 = ?
To find the molarity of the phosphoric acid solution, you need to use the balanced chemical equation and the volume and molarity of the potassium hydroxide (KOH) solution.
The balanced chemical equation for the neutralization reaction is:
H3PO4 + 3 KOH -> K3PO4 + 3 H2O
From the equation, you can see that the ratio between H3PO4 and KOH is 1:3. This means that 1 mole of H3PO4 reacts with 3 moles of KOH.
You are given the volume and molarity of the KOH solution. The volume of KOH solution is 20.0 mL, and the molarity is 1.200 M.
To find the moles of KOH used, you can use the formula:
moles of KOH = volume (L) x molarity (mol/L)
First, convert the volume of KOH solution from mL to L:
20.0 mL = 20.0 mL / 1000 mL/L = 0.020 L
Now, calculate the moles of KOH used:
moles of KOH = 0.020 L x 1.200 mol/L = 0.024 mol
Since the ratio between H3PO4 and KOH is 1:3, there must be 0.024 moles of H3PO4.
Now, you are provided with the volume of the H3PO4 solution, which is 60.0 mL. To find the molarity of the H3PO4 solution, you need to convert the volume to L and divide the moles by the volume:
60.0 mL = 60.0 mL / 1000 mL/L = 0.060 L
Molarity of H3PO4 solution = moles of H3PO4 / volume of H3PO4 solution
= 0.024 mol / 0.060 L
= 0.400 M
Therefore, the molarity of the phosphoric acid solution is 0.400 M.