Answer :
Final answer:
To find the volume of 1.420 M NaOH needed to titrate 44.50 mL of 1.500 M H3PO4, we use stoichiometry to calculate the moles required and then determine the volume needed. The final calculated volume is 141 mL, rounded to three significant digits.
Explanation:
To calculate the volume of a 1.420 M NaOH solution required to titrate 44.50 mL of a 1.500 M H3PO4 solution, we need to know the stoichiometry of the reaction between NaOH and H3PO4. Each molecule of H3PO4 requires three molecules of NaOH to completely react:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
Let's calculate the number of moles of H3PO4 to be titrated:
(1.500 mol/L) × (0.0445 L) = 0.06675 moles H3PO4
Since the stoichiometry is 1:3, we need three times as many moles of NaOH:
0.06675 moles H3PO4 × 3 moles NaOH / 1 mole H3PO4 = 0.20025 moles NaOH
We can use the molarity of the NaOH solution to find the volume required:
Volume = moles / molarity = 0.20025 moles NaOH / 1.420 mol/L = 0.14105634... L
Converting liters to milliliters (1 L = 1000 mL) gives us the volume in milliliters:
0.14105634... L × 1000 mL/L = 141.05634... mL
To match the significant digits of the least precise measurement used in our calculations, we round our result to three significant digits:
141 mL