Suppose that 50.0mL of 0.240M sulfric acid requires 36.0mL of KOH solution to reach its endpoint. What is the molarity of the KOH solution?
This works the same way as the Ba(OH)2 and HCl problem.
The equation is H2SO4 + 2KOH ==> K2SO4 = 2H2O
0.667
To find the molarity of the KOH solution, we can use the concept of stoichiometry and the balanced chemical equation of the reaction between sulfuric acid (H2SO4) and potassium hydroxide (KOH).
The balanced chemical equation for this reaction is:
H2SO4 + 2 KOH -> K2SO4 + 2 H2O
From the balanced equation, we can see that 1 mole of sulfuric acid reacts with 2 moles of KOH.
Given that 50.0 mL of 0.240 M sulfuric acid reacts with 36.0 mL of KOH solution, we can convert the volume of sulfuric acid into moles by using its molarity:
moles of H2SO4 = volume (L) x molarity (mol/L)
= 50.0 mL x (1 L / 1000 mL) x 0.240 mol/L
= 0.012 mol
Using the mole ratio from the balanced equation, we can determine the number of moles of KOH that reacted:
moles of KOH = moles of H2SO4 x (2 moles KOH / 1 mole H2SO4)
= 0.012 mol x (2 mol / 1 mol)
= 0.024 mol
Finally, we can calculate the molarity of the KOH solution by dividing the number of moles by the volume in liters:
molarity of KOH = moles of KOH / volume of KOH (L)
= 0.024 mol / (36.0 mL x (1 L / 1000 mL))
= 0.667 M
Therefore, the molarity of the KOH solution is 0.667 M.