What volume in ml of 0.0985 m NaOH solution is required to reach the end point in the complete titration of a 15.0 ml sample of 0.124 M phosphoric acid?
H3PO4 + 3NaOH ==> Na3PO4 + 3H2O
mols H3PO4 = M x L = ?
mols NaOH = 3x that (from the coefficients in the balanced equation).
M NaOH = mols NaOH/L NaOH. You know M and mols, solve for L and change to mL.
To determine the volume of the NaOH solution required to reach the end point in the titration, we can use the concept of stoichiometry. Here's how you can calculate it:
Step 1: Write the balanced chemical equation for the reaction between NaOH and phosphoric acid (H3PO4):
H3PO4 + 3NaOH -> Na3PO4 + 3H2O
Step 2: Identify the stoichiometric ratio between NaOH and H3PO4 from the balanced equation. In this case, it is 1:3. This means that one mole of phosphoric acid reacts with three moles of sodium hydroxide.
Step 3: Calculate the number of moles of phosphoric acid in the 15.0 ml sample of 0.124 M phosphoric acid. Molarity (M) is defined as moles of solute divided by the volume of solution in liters:
moles of H3PO4 = Molarity × volume (in liters)
moles of H3PO4 = 0.124 mol/L × (15.0 mL / 1000 mL/L)
Step 4: Determine the volume of 0.0985 M NaOH solution required to neutralize the moles of phosphoric acid. Since the stoichiometric ratio is 1:3, we need three times the number of moles of NaOH:
moles of NaOH = 3 × moles of H3PO4
Step 5: Calculate the volume of NaOH solution using its molarity and the number of moles:
volume (in liters) = moles of NaOH / Molarity
volume (in mL) = volume (in liters) × 1000
Now, plug in the values and perform the calculations:
moles of H3PO4 = 0.124 mol/L × (15.0 mL / 1000 mL/L) = 0.00186 moles
moles of NaOH = 3 × 0.00186 = 0.00558 moles
volume (in liters) = 0.00558 moles / 0.0985 mol/L = 0.0566 L
volume (in mL) = 0.0566 L × 1000 = 56.6 mL
Therefore, approximately 56.6 mL of 0.0985 M NaOH solution is required to reach the end point in the complete titration of a 15.0 mL sample of 0.124 M phosphoric acid.