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?
To find the molarity of the KOH solution, we can use the equation:
Molarity(KOH) x Volume(KOH) = Molarity(H2SO4) x Volume(H2SO4)
Given:
Volume(KOH) = 40.0 mL
Molarity(H2SO4) = 1.50 M
Volume(H2SO4) = 15.7 mL
Plugging in the values, we get:
Molarity(KOH) x 40.0 mL = 1.50 M x 15.7 mL
Now solve for Molarity(KOH):
Molarity(KOH) = (1.50 M x 15.7 mL) / 40.0 mL
Molarity(KOH) = 0.5895 M
Therefore, the molarity of the KOH solution is 0.5895 M.