Thank you for visiting Calculate the volume of a 1 420 M NaOH solution required to titrate 44 50 mL of a 1 500 M tex H 3PO 4. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
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
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