A volume of 40.0 of aqueous potassium hydroxide was titrated against a standard solution of sulfuric acid h2so4. What was the molarity of the solution if 20.02 of 1.50M h2s04 was needed?
Who helps me to solve it for me.
Technically the problem can't be solved because you have no units for 40.0 and 20.02. Assuming those are mL, the problem is solved as follows:
2KOH + H2SO4 ==> K2SO4 + 2H2O
mols H2SO4 = M x L = ?
mols KOH = 2x that (look at the equation coefficients)
Then M KOH = mols KOH/L KOH.
To solve this problem, you can use the concept of stoichiometry and the equation for the reaction between potassium hydroxide (KOH) and sulfuric acid (H2SO4).
The balanced chemical equation for the reaction is:
2 KOH + H2SO4 → K2SO4 + 2 H2O
In this reaction, it takes 2 moles of KOH to react with 1 mole of H2SO4. Therefore, to determine the molarity of the potassium hydroxide (KOH) solution, we can use the formula:
Molarity of KOH solution = (moles of H2SO4) / (volume of KOH solution in liters)
First, calculate the moles of H2SO4 used in the titration. Given that 20.02 mL of 1.50M H2SO4 was needed, we can convert the volume to liters:
20.02 mL = 20.02 / 1000 L = 0.02002 L
Next, determine the moles of H2SO4:
moles of H2SO4 = (Molarity of H2SO4) x (volume of H2SO4 in liters)
= (1.50 mol/L) x (0.02002 L)
Now, we can use the stoichiometry of the balanced equation to find the moles of KOH used:
moles of KOH = (moles of H2SO4) x (2 moles of KOH / 1 mole of H2SO4)
Finally, we can determine the molarity of the KOH solution by dividing the moles of KOH by the volume of the KOH solution in liters (which is given as 40.0 mL):
Molarity of KOH solution = (moles of KOH) / (volume of KOH solution in liters)
= (moles of KOH) / (40.0 mL / 1000 L)
By performing these calculations, you will be able to determine the molarity of the KOH solution used in the titration.