A 10.0mL sample of H2SO4 was neutralized by 13.5mL of 1.0M KOH. What was its molarity of H2SO4? What is the normality?

Normality*volume=normality*volume

Normalityacid=1*13.5/10=13.5

Molarityacid= 13.5/2

Well, it sounds like we're dealing with some chemistry here. Don't worry, I won't get too "acidic" with my humor!

To find the molarity of H2SO4, we can use the equation:

M1 * V1 = M2 * V2

Where:
M1 = molarity of the first solution (H2SO4)
V1 = volume of the first solution (10.0 mL)
M2 = molarity of the second solution (KOH)
V2 = volume of the second solution (13.5 mL)

Since we want to find the molarity of H2SO4, let's plug in the values:

M1 * 10.0 mL = 1.0 M * 13.5 mL

Now, let's solve for M1:

M1 = (1.0 M * 13.5 mL) / 10.0 mL

M1 = 13.5 M/mL

So, the molarity of H2SO4 is 13.5 M/mL. Woah, that's quite concentrated!

Now, for the normality. The normality is the number of equivalents of solute per liter of solution. In this case, H2SO4 is a diprotic acid, which means it can donate two protons (two equivalents) per molecule.

Since we used 10.0 mL of H2SO4 and its molarity is 13.5 M/mL, we can calculate the number of moles of H2SO4:

moles of H2SO4 = Molarity * Volume (in liters) = 13.5 M/mL * 0.010 L

Now, since H2SO4 is a diprotic acid, we have twice the amount of moles of protons. Therefore, the number of moles of protons is:

moles of protons = 2 * moles of H2SO4

With the number of moles of protons, we can find the normality:

Normality = moles of protons / volume (in liters)

Now that's quite a "normal"ity calculation. But hey, it's just chemistry trying to make our brains do somersaults!

To determine the molarity of H2SO4, we can use the stoichiometry of the neutralization reaction. The balanced equation for the reaction between H2SO4 and KOH is:

H2SO4 + 2KOH -> K2SO4 + 2H2O

From the equation, we see that the mole ratio between H2SO4 and KOH is 1:2. This means that 1 mole of H2SO4 reacts with 2 moles of KOH.

First, let's calculate the number of moles of KOH used:

volume of KOH = 13.5 mL = 0.0135 L
molarity of KOH = 1.0 M

moles of KOH = volume of KOH x molarity of KOH
= 0.0135 L x 1.0 M
= 0.0135 moles

Since the mole ratio between H2SO4 and KOH is 1:2, the number of moles of H2SO4 is half of the moles of KOH:

moles of H2SO4 = 0.0135 moles / 2
= 0.00675 moles

Next, we need to calculate the volume of H2SO4 in liters:

volume of H2SO4 = 10.0 mL = 0.010 L

Finally, molarity of H2SO4 can be calculated using the formula:

molarity = moles / volume

molarity of H2SO4 = 0.00675 moles / 0.010 L
= 0.675 M

The molarity of H2SO4 is 0.675 M.

Now, let's calculate the normality of H2SO4. Normality is a measure of the number of reactive species (in this case, H+ ions) in a solution.

For H2SO4, there are 2 moles of H+ ions per mole of H2SO4. Therefore, the normality of H2SO4 is twice its molarity:

normality of H2SO4 = 2 x molarity of H2SO4
= 2 x 0.675 M
= 1.35 N

The normality of H2SO4 is 1.35 N.

To determine the molarity of H2SO4, we can use the equation:

Molarity of Acid × Volume of Acid = Molarity of Base × Volume of Base

In this case, the volume of acid (H2SO4) is 10.0 mL and the volume of base (KOH) is 13.5 mL. The molarity of the base is given as 1.0 M.

Plugging in these values into the equation:

Molarity of Acid × 10.0 mL = 1.0 M × 13.5 mL

Rearranging the equation to solve for the molarity of the acid:

Molarity of Acid = (1.0 M × 13.5 mL) / 10.0 mL

Molarity of Acid = 13.5 M/mL / 10.0 mL

Molarity of Acid = 1.35 M

Therefore, the molarity of H2SO4 is 1.35 M.

To determine the normality, we need to consider the number of acidic hydrogen atoms that can react with 1 mole of base. In the case of H2SO4, it is a diprotic acid, which means it can donate two hydrogen ions (H+) per molecule.

Since 1 mole of H2SO4 can donate 2 moles of H+, the normality (N) is twice the molarity (M) of H2SO4.

Normality = 2 × Molarity of Acid

Normality = 2 × 1.35 M

Normality = 2.7 N

Therefore, the normality of H2SO4 is 2.7 N.