A volume of 50.0 mL of aqueous potassium hydroxide was titrated against a standard solution of sulfuric acid. What was the molarity of the KOH solution if 19.2 mL of 1.50 M H2SO4 was needed? The equation is
2KOH(aq) +H2SO4(aq)-» K2SO4(aq)+2H2O(l)
To find the molarity (M) of the KOH solution, we can use the stoichiometry of the balanced equation and the volume and molarity of the H2SO4 solution used.
First, let's identify the stoichiometric ratio between KOH and H2SO4 from the balanced equation:
2KOH(aq) + H2SO4(aq) -> K2SO4(aq) + 2H2O(l)
According to the equation, 2 moles of KOH react with 1 mole of H2SO4. This means that the mole ratio of KOH to H2SO4 is 2:1.
Now we can set up a proportion using the given information:
(1.50 mol H2SO4 / 1 L) * (19.2 mL H2SO4 / 1000 mL) = (x mol KOH / 50.0 mL KOH)
Note that we convert mL to L by dividing by 1000 since 1 L = 1000 mL.
Simplifying the equation, we have:
1.50 * 19.2 / 1000 = x / 50.0
Now we can solve for x (the number of moles of KOH):
x = (1.50 * 19.2 * 50.0) / 1000
x ≈ 1.44 moles KOH
Finally, to find the molarity of the KOH solution, we use the formula:
Molarity (M) = Moles of solute / Volume of solution in liters
Molarity (KOH) = 1.44 moles / 0.050 L
Molarity (KOH) ≈ 28.8 M
moles H2SO4 = M x L.
From the equation, moles KOH must be twice moles H2SO4.
M KOH = moles/L.