76.5 mL of 2.50 M sulfuric acid are required to titrate a sample of sodium hydroxide to the endpoint. If the volume of the sample of NaOH is 35.00 mL , what is the molarity of the NaOH solution?H2SO4 (aq) + 2NaOH (aq) ⟶ Na2SO4 (aq) + 2H2O (l)
Question
76.5 mL of 2.50 M sulfuric acid are required to titrate a sample of sodium hydroxide to the endpoint. If the volume of the sample of NaOH is 35.00 mL , what is the molarity of the NaOH solution?H2SO4 (aq) + 2NaOH (aq) ⟶ Na2SO4 (aq) + 2H2O (l)
Solution
Sure, here is the step by step solution:
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First, we need to find the moles of H2SO4. We can do this by multiplying the volume (in liters) by the molarity.
Moles of H2SO4 = Molarity * Volume = 2.50 mol/L * 0.0765 L = 0.19125 mol
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From the balanced chemical equation, we know that the ratio of H2SO4 to NaOH is 1:2. Therefore, the moles of NaOH is twice the moles of H2SO4.
Moles of NaOH = 2 * Moles of H2SO4 = 2 * 0.19125 mol = 0.3825 mol
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Finally, we can find the molarity of the NaOH solution by dividing the moles of NaOH by the volume (in liters).
Molarity of NaOH = Moles of NaOH / Volume = 0.3825 mol / 0.035 L = 10.93 M
So, the molarity of the NaOH solution is 10.93 M.
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