28.5 mL of 3.00 M sulfuric acid are required to titrate a sample of sodium hydroxide to the endpoint. If the volume of the sample of NaOH is 25.00 mL , what is the molarity of the NaOH solution?H2SO4 (aq) + 2NaOH (aq) ⟶ Na2SO4 (aq) + 2H2O (l)
Question
28.5 mL of 3.00 M sulfuric acid are required to titrate a sample of sodium hydroxide to the endpoint. If the volume of the sample of NaOH is 25.00 mL , what is the molarity of the NaOH solution?H2SO4 (aq) + 2NaOH (aq) ⟶ Na2SO4 (aq) + 2H2O (l)
Solution
Sure, here are the steps to solve this problem:
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First, we need to find out how many moles of H2SO4 are present. We can do this by multiplying the volume of the H2SO4 solution by its molarity.
Moles of H2SO4 = Volume (L) x Molarity (M) Moles of H2SO4 = 0.0285 L x 3.00 M = 0.0855 moles
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From the balanced chemical equation, we know that one mole of H2SO4 reacts with two moles of NaOH. Therefore, the moles of NaOH in our sample is twice the moles of H2SO4.
Moles of NaOH = 2 x Moles of H2SO4 Moles of NaOH = 2 x 0.0855 moles = 0.171 moles
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Finally, we can find the molarity of the NaOH solution by dividing the moles of NaOH by the volume of the NaOH solution.
Molarity of NaOH = Moles of NaOH / Volume (L) Molarity of NaOH = 0.171 moles / 0.025 L = 6.84 M
So, the molarity of the NaOH solution is 6.84 M.
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