In a silphuric acid (H2SO4)-Sodium Hydroxide (NaOH) acid-base titration, 17.3mL of 0.126M NaOH is needed to neutralize 25mL of H2SO4 of unknown concentration. What is the morality of the H2SO4 solution?
H2SO4 (aq)+ NaOH (aq) = Na2SO4 (aq) + H2O (l)
morality? Most H2SO4 solutions are not moral at all. Neither are NaOH solution.
H2SO4 + 2NaOH ==> Na2SO4 + 2H2O
mols NaOH = M x L = ?
mols H2SO4 = 1/2 mols NaOH which you get from the coefficients in the balanced equation.
Then M H2SO4 = mols H2SO4/L H2SO4
M is for molarity.
40.8
To determine the molarity of the H2SO4 solution, we can use the balanced chemical equation for the reaction. From the equation, we know that 1 mole of H2SO4 reacts with 2 moles of NaOH to form 1 mole of Na2SO4 and 2 moles of H2O.
Given:
Volume of NaOH solution (V1) = 17.3 mL = 0.0173 L
Molarity of NaOH solution (M1) = 0.126 M
Volume of H2SO4 solution (V2) = 25 mL = 0.025 L
The balanced equation tells us that the ratio of moles of NaOH to H2SO4 is 2:1. Therefore, we can set up a proportion to find the moles of H2SO4:
(0.126 moles NaOH / 1 L) * (0.0173 L) = (x moles H2SO4 / 0.025 L)
Solving for x, we find the moles of H2SO4 in the 25 mL of solution. Next, we can calculate the molarity of the H2SO4 solution using the formula:
Molarity (M) = moles of solute / volume of solution (in liters)
Substituting the known values, we have:
Molarity (H2SO4) = x moles H2SO4 / 0.025 L
Now, plug in the value of x from the previous calculation and solve for the molarity of the H2SO4 solution.