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?
poopy
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
mols H2SO4 = M x L = ?
Convert mols H2SO4 to mols NaOH using the coefficients in the balanced equation.
Then M NaOH = mols NaOH/L NaOH.
To find the molarity of the sodium hydroxide solution, we can use the equation:
M1V1 = M2V2
Where:
M1 = molarity of sulfuric acid (known as 0.75M)
V1 = volume of sulfuric acid used (known as 14 mL)
M2 = molarity of sodium hydroxide (to be calculated)
V2 = volume of sodium hydroxide used (known as 38 mL)
Substituting the known values into the equation, we have:
(0.75M)(14 mL) = M2(38 mL)
Rearranging the equation to solve for M2 (molarity of sodium hydroxide):
M2 = (0.75M)(14 mL) / (38 mL)
Now, let's calculate the molarity of the sodium hydroxide solution using this equation:
M2 = (0.75M)(14 mL) / (38 mL)
M2 = 0.3125 M
Therefore, the molarity of the sodium hydroxide solution is 0.3125 M.
To find the molarity of the sodium hydroxide solution, we can use the concept of stoichiometry and the balanced chemical equation between sodium hydroxide (NaOH) and sulfuric acid (H2SO4). The balanced equation is:
2NaOH + H2SO4 -> Na2SO4 + 2H2O
From the balanced equation, we can see that it takes 2 moles of sodium hydroxide to react with 1 mole of sulfuric acid. This means that the ratio of moles of sodium hydroxide to moles of sulfuric acid is 2:1.
Given that the volume of the sulfuric acid used is 14 mL and its concentration is 0.75M, we can calculate the moles of sulfuric acid used as follows:
Moles of H2SO4 = Molarity × Volume (in liters)
= 0.75 mol/L × 0.014 L
≈ 0.0105 mol
Since the stoichiometric ratio of NaOH to H2SO4 is 2:1, we know that the moles of sodium hydroxide used in the reaction are twice the moles of sulfuric acid used:
Moles of NaOH = 2 × Moles of H2SO4
= 2 × 0.0105 mol
= 0.021 mol
Now, to find the molarity of the sodium hydroxide solution, we need to divide the moles of sodium hydroxide by the volume of the solution used in the titration. The volume of the sodium hydroxide solution used is 38 mL, but it needs to be converted to liters:
Volume of NaOH solution (in L) = 38 mL ÷ 1000 mL/L
= 0.038 L
Finally, we can calculate the molarity of the sodium hydroxide solution:
Molarity of NaOH = Moles of NaOH ÷ Volume of NaOH solution (in L)
= 0.021 mol ÷ 0.038 L
≈ 0.553 M
Therefore, the molarity of the sodium hydroxide solution is approximately 0.553 M.