A volume of 60.0 of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 25.2 mL of 1.50M H2SO4 was needed? The equation is 2KOH+H2SO4-->K2SO4+2H2O. I cant seem to get the right answer so please, if you know how to do this explain it fully and work out the problem completely with numbers and equations.

Question

A volume of 60.0 of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 25.2 mL of 1.50M H2SO4 was needed? The equation is 2KOH+H2SO4-->K2SO4+2H2O. I cant seem to get the right answer so please, if you know how to do this explain it fully and work out the problem completely with numbers and equations.

Answer 1

Has it occurred to you that if you had posted your work I could have found the error in half the time it takes to work the problem.

H2SO4 + 2KOH ==> K2SO4 + 2H2O

How many moles H2SO4 were used? That is M x L = moles H2SO4.
Now convert that to moles KOH. From the equation, it takes 2 moles KOH to equal 1 mole H2SO4; therefore, moles H2SO4 x 1/2 = moles KOH.
Now M KOH = moles KOH/L KOH.

Answer 2

To find the molarity of the aqueous KOH solution, we can use the following equation:

2KOH + H2SO4 -> K2SO4 + 2H2O

First, let's write the balanced chemical equation:

2 moles of KOH + 1 mole of H2SO4 -> 1 mole of K2SO4 + 2 moles of H2O

From the balanced equation, we can see that the stoichiometric ratio between KOH and H2SO4 is 2:1. This means that for every 1 mole of H2SO4, we need 2 moles of KOH to react with it.

Given that 25.2 mL of 1.50 M H2SO4 was needed, we can start by calculating the number of moles of H2SO4 used:

Moles of H2SO4 = (volume of H2SO4 in L) x (molarity of H2SO4)
= (25.2 mL / 1000 mL/L) x (1.50 mol/L)
= 0.0378 mol

Since the stoichiometric ratio between KOH and H2SO4 is 2:1, the number of moles of KOH used in the reaction is half of the moles of H2SO4 used:

Moles of KOH = (1/2) x moles of H2SO4
= (1/2) x 0.0378 mol
= 0.0189 mol

Now, we can calculate the volume of the aqueous KOH solution using its Molarity and the moles of KOH used:

Molarity of KOH = (moles of KOH) / (volume of KOH in L)

Rearranging the equation, we get:

Volume of KOH in L = (moles of KOH) / (Molarity of KOH)

Since the volume of KOH solution used is given as 60.0 mL, we need to convert it into liters:

Volume of KOH in L = 60.0 mL / 1000 mL/L
= 0.0600 L

Now we can substitute the values into the equation to find the Molarity of KOH:

Molarity of KOH = (0.0189 mol) / (0.0600 L)
= 0.315 M

Therefore, the molarity of the KOH solution is 0.315 M.