A volume of 40.0 mL
of aqueous potassium hydroxide ( KOH
) was titrated against a standard solution of sulfuric acid ( H2SO4
). What was the molarity of the KOH
solution if 15.7 mL
of 1.50 M
H2SO4
was needed? The equation is
2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)
To solve this problem, we first need to determine the number of moles of sulfuric acid used in the titration. We can do this using the formula:
Moles of H2SO4 = Molarity of H2SO4 x Volume of H2SO4
Moles of H2SO4 = 1.50 M x 15.7 mL = 23.55 mmol
Since the balanced chemical equation shows that 2 moles of KOH react with 1 mole of H2SO4, we can determine the number of moles of KOH present in the solution:
Moles of KOH = Moles of H2SO4 / 2 = 23.55 mmol / 2 = 11.775 mmol
Now, to calculate the molarity of the KOH solution, we use the formula:
Molarity of KOH = Moles of KOH / Volume of KOH (in L)
Volume of KOH = 40.0 mL = 0.040 L
Molarity of KOH = 11.775 mmol / 0.040 L = 293.75 mmol/L = 0.2937 M
Therefore, the molarity of the KOH solution is 0.2937 M.