By titration it is found that 49.3 ml of 0.151 m naoh(aq) is needed to neutralize 25.0 ml of hcl(aq). calculate the concentration of the hcl solution.

0.298 M The balanced equation for the reaction is HCl + NaOH = H2O + NaCl So for every mole of NaOH used, one mole of HCl is neutralized. So let's determine how many moles of NaOH we used. That will simply be the volume (in liters) of solution multiplied by the molarity. So 0.0493 L * 0.151 mol/L = 0.0074443 mol Because of that 1:1 ratio, we now know that 0.0074443 moles of HCl was in the sample. Since molarity is moles per liter, a little division will give us the molarity of the HCl solution. So 0.0074443 mol / 0.025 L = 0.297772 mol/M = 0.297772 M Rounding to 3 significant figures gives 0.298 M.

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