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How many moles of hydrochloric acid are present in a solution of 50 mL of 0.1M NaOH when titrated with 0.1M HCl solution?

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To find the moles of hydrochloric acid  HCl  present in the solution when titrated with 0.1M NaOH, we can use the concept of stoichiometry. The balanced chemical equation for the reaction between NaOH and HCl is:NaOH + HCl  NaCl + H2OFrom the balanced equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl. First, we need to find the moles of NaOH in the 50 mL solution:moles of NaOH = Molarity  Volume  in liters moles of NaOH = 0.1 M   50 mL  0.001 L/mL moles of NaOH = 0.1 M  0.05 Lmoles of NaOH = 0.005 molesSince the stoichiometry between NaOH and HCl is 1:1, the moles of HCl present in the solution will be equal to the moles of NaOH:moles of HCl = 0.005 molesSo, there are 0.005 moles of hydrochloric acid present in the solution when titrated with 0.1M NaOH.

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