How many mL of 0.630M NaOH would be needed to react with 20.0mL of a 0.400M solution of H3PO4? H3PO4(aq)+3NaOH(aq)-->3H2O(l)+Na3PO4

Answer 1

Based on the equation given, it will take 3 times as many mols of NaOH to fully neutralize the H3PO4. This is determined by the numbers in front of the molecular formulas on the left side of the equation.

With that in mind, the best way to start this problem is by calculating the mols of H3PO4.

1. ) Since M = mol/L, convert 20 mL to 0.020 L by moving the decimal 3 places to the left.
2. ) Multiply 0.020 L x 0.400 mol/L to cancel the Liters. This equals 0.008 mols of H3PO4 to neutralize.

Next, multiply 0.008 mols by 3 to find the number of mols of NaOH you need = 0.24 mols of NaOH.

This reduces the initial question to how many mL of 0.630 M NaOH gives you 0.024 mols of NaOH.

So:

Now multiply 0.024 mols x 1 L/0.630 mols = 0.0381 L of NaOH.

The reason you need to multiply by 1/0.630 (which is the inverse of 0.630 M) is because the mols have to be on the bottom to cancel. This gives you the volume in Liters.

Converting 0.0381 L to mL gives you a final answer of 38.1 L of NaOH.