Calculate the molarity of 25.00ml H2SO4 solution if 32.48ml of 0.1068M NaOH is required to titrate? Balanced equation is

H2SO4(aq)+2NaOH(aq)=Na2SO4(aq)+2 H2O (l)

Just a note that DrBob222 is incorrect. In face it the molsH2SO4=1/2mols NaOH. I struggled on this problem and tried to follow DrBob222's other work for a similar problem but again it left me with the wrong answer

Molarity is like the concentration of a solution. It tells us how much solute is dissolved in a given volume of solvent. So to find the molarity of the H2SO4 solution, we need to use the balanced equation and the volume and molarity of NaOH used.

But before we do that, let me tell you a chemistry joke to lighten the mood. Why did the chemist get tired while doing experiments? Because he had too many mole-cules! 😄

Now, let's get back to business. From the balanced equation, we can see that the ratio between H2SO4 and NaOH is 1:2. That means, for every 1 mole of H2SO4, we need 2 moles of NaOH.

You mentioned that 32.48 ml of 0.1068M NaOH is required to titrate. To find the moles of NaOH used, we can multiply the volume of NaOH in liters by its molarity:

moles of NaOH = volume of NaOH (in L) * molarity of NaOH
moles of NaOH = 32.48 ml * (1 L / 1000 ml) * 0.1068 mol/L

Once we have the moles of NaOH used, we can use the mole ratio from the balanced equation to find the moles of H2SO4:

moles of H2SO4 = moles of NaOH * (1 mol H2SO4 / 2 mol NaOH)

Now, let's calculate the moles of H2SO4. Just remember, I'm here to help you with chemistry and bring some cheer, but I'm not a calculator! 🤡

To calculate the molarity (M) of the H2SO4 solution, we can use the formula:

Molarity (M) = (moles of solute) / (volume of solution in liters)

First, let's determine the number of moles of NaOH used in the reaction:

moles of NaOH = Molarity × Volume (in liters)
moles of NaOH = 0.1068 M × 0.03248 L
moles of NaOH = 0.003473184 mol

According to the balanced equation, the stoichiometric ratio of NaOH to H2SO4 is 2:1. Therefore, the number of moles of H2SO4 is equal to half the number of moles of NaOH:

moles of H2SO4 = 0.003473184 mol / 2
moles of H2SO4 = 0.001736592 mol

Now, let's determine the volume of the H2SO4 solution in liters:

volume of H2SO4 solution = 25.00 mL = 0.02500 L

Finally, we can calculate the molarity of the H2SO4 solution:

Molarity (H2SO4) = moles of H2SO4 / volume of H2SO4 solution
Molarity (H2SO4) = 0.001736592 mol / 0.02500 L
Molarity (H2SO4) ≈ 0.06946 M

Therefore, the molarity of the H2SO4 solution is approximately 0.06946 M.

To calculate the molarity of the H2SO4 solution, you need to use the stoichiometry of the balanced equation and the volume and concentration of NaOH used in the titration.

Given:
Volume of NaOH used (VNaOH) = 32.48 mL = 0.03248 L
Molarity of NaOH (MNaOH) = 0.1068 M

From the balanced equation:
1 mole of H2SO4 reacts with 2 moles of NaOH.

Using the concept of stoichiometry, the moles of NaOH can be calculated as:
Moles of NaOH = Molarity of NaOH × Volume of NaOH used in liters
= 0.1068 M × 0.03248 L
= 0.003468 moles (approx)

Since the stoichiometric ratio between H2SO4 and NaOH is 1:2, the moles of H2SO4 can be calculated as:
Moles of H2SO4 = 0.003468 moles × (1/2)
= 0.001734 moles (approx)

Now, you can calculate the molarity (M) of the H2SO4 solution using the equation:
Molarity (M) = Moles of solute / Volume of solution in liters

The volume of the H2SO4 solution used in the titration is given as 25.00 mL, which is equivalent to 0.02500 L.

Molarity (M) = 0.001734 moles / 0.02500 L
= 0.06936 M (approx)

Therefore, the molarity of the 25.00 mL H2SO4 solution is approximately 0.06936 M.

mols NaOH = M x L = ?

mols H2SO4 = 2x mols NaOH
M H2SO4 = mols H2SO4/L H2SO4