Calculate the molarity of the H2SO4 solution if 17.45 mL of NaOH was necessary to reach the endpoint of titration. The molarity of the NaOH solution was 0.425 M and 26.30 mL of H2SO4 was added to the Erlenmeyer flask.
H2SO4(aq) + 2 NaOH (aq) -----> Na2SO4 (aq) + 2 H2O (l)
I'm not sure how to set this problem correctly, thanks for the help.
hydrogen and hydroxide ions are neutralizing each other
acid vol * molarity * hydrogens =
... base vol * molarity * hydroxides
26.3 * x * 2 = 17.45 * .425 * 1
To calculate the molarity of the H2SO4 solution, we need to use the balanced chemical equation and stoichiometry. The balanced equation shows the relationship between the reactants and products in terms of moles.
First, let's determine the number of moles of NaOH used in the reaction. We can use the formula:
moles of NaOH = volume (in liters) x concentration (in moles per liter)
Given that the volume of NaOH used is 17.45 mL (which is equivalent to 0.01745 L) and the concentration of NaOH is 0.425 M, we can calculate the moles of NaOH used:
moles of NaOH = 0.01745 L x 0.425 M = 0.00741625 moles
According to the balanced equation, the stoichiometric ratio between NaOH and H2SO4 is 1:1. This means that for every mole of NaOH used, there is an equivalent mole of H2SO4. Therefore, we have 0.00741625 moles of H2SO4 in the reaction.
Next, we need to determine the volume of H2SO4 added to the Erlenmeyer flask. The volume given is 26.30 mL (which is equivalent to 0.02630 L).
To calculate the molarity of the H2SO4 solution, we divide the moles of H2SO4 by the volume of H2SO4 in liters:
molarity of H2SO4 = moles of H2SO4 / volume of H2SO4 (in liters)
molarity of H2SO4 = 0.00741625 moles / 0.02630 L = 0.282 M
Therefore, the molarity of the H2SO4 solution is 0.282 M.