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 .75M sulfuric acid?
To calculate the molarity of the sodium hydroxide solution, you need to use the equation of the reaction that occurs during the titration. In this case, sodium hydroxide (NaOH) reacts with sulfuric acid (H2SO4) in a 1:2 ratio.
The balanced chemical equation for the reaction is as follows:
2NaOH + H2SO4 → Na2SO4 + 2H2O
From the balanced equation, we can determine that 2 moles of NaOH react with 1 mole of H2SO4. This means that the number of moles of NaOH can be calculated from the moles of H2SO4 used in the titration.
To find the moles of H2SO4, we can use the formula:
moles = molarity × volume (in liters)
Given that the volume of sulfuric acid used is 14 mL and the molarity of the sulfuric acid is 0.75 M, we can calculate the moles of H2SO4 as follows:
moles of H2SO4 = 0.75 M × 0.014 L = 0.0105 moles
Since the reaction ratio is 2:1, the moles of NaOH will be twice the moles of H2SO4 used:
moles of NaOH = 2 × 0.0105 moles = 0.021 moles
Now, to determine the molarity of the sodium hydroxide solution, we need to divide the moles of NaOH by the volume of the solution used.
The volume of the sodium hydroxide solution used is given as 38 mL. However, it is important to note that the solution volume should be converted to liters to maintain consistency with the units of molarity.
Converting 38 mL to liters:
38 mL ÷ 1000 = 0.038 L
Finally, we can calculate the molarity of the sodium hydroxide solution:
Molarity (M) = moles ÷ volume (in liters)
Molarity = 0.021 moles ÷ 0.038 L = 0.553 M
Therefore, the molarity of the sodium hydroxide solution is 0.553 M.