What is the molarity of a sodium hydroxide solution if 38 mL of the solution is titrated to the end point with 14 mL of 0.75M sulfuric acid?
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To find the molarity of the sodium hydroxide (NaOH) solution, we can use the principle of stoichiometry. In this case, we know that sulfuric acid (H2SO4) reacts with sodium hydroxide in a 1:2 ratio.
First, let's calculate the number of moles of sulfuric acid used in the reaction:
Moles of sulfuric acid = molarity × volume
= 0.75 M × 0.014 L
= 0.0105 moles
According to the stoichiometric ratio, for every mole of sulfuric acid, two moles of NaOH are required. Therefore, the number of moles of NaOH in the solution is twice the number of moles of sulfuric acid used:
Moles of NaOH = 2 × 0.0105 moles
= 0.021 moles
Next, we need to calculate the volume of the NaOH solution used in the titration. Since all the reactants are in the liquid state, we can assume that the volumes of the solutions are directly proportional to the number of moles present:
Volume of NaOH solution = (volume of H2SO4 solution × moles of NaOH) / moles of H2SO4
= (0.038 L × 0.021 moles) / 0.0105 moles
= 0.076 L (or 76 mL)
Finally, we can calculate the molarity of the NaOH solution:
Molarity of NaOH solution = moles of NaOH / volume of NaOH solution
= 0.021 moles / 0.076 L
≈ 0.276 M
Therefore, the molarity of the sodium hydroxide solution is approximately 0.276 M.