25.mL of HCl was titrated against NaOH. It required 42mL of .10M NaOH to neutralize 25mL of acid. How many moles of acid were in the sample? HCl+NaOH --> H2O+NaCl,
First we look at the balanced chemical equation
HCl + NaOH --> H2O + NaCl
as well as the information given about each substance in the problem. Since we have a balanced chemical equation, we can see that the mole ratio between HCl and NaOH is 1:1. For every one mole of NaOH added, we must have had 1 mol of HCl in the solution.
We can set this problem up as a stoichiometry problem.
42 mL NaOH (1 L/1000 mL)(0.10 mol NaOH /L) (1 mol HCl / 1 mol NaOH) =
0.0042 moles of HCl in the 25 mL sample.
If you should need to, you can find the molarity of the HCl sample by taking the moles of HCl and dividing by the volume of HCl in L (0.025 L) to find
0.0042 mol / 0.025 = 0.168 M HCl
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