A 25.0-mL sample of H2SO4 is neutralized with NaOH. What is the concentration of the H2SO4

if 35.0 mL of 0.150 M NaOH are required to
completely neutralize the acid?

.105 M

0.180M

There seems to be a mistake in the given information. We can calculate the concentration of H2SO4 using the given information as follows:

NaOH and H2SO4 react in a 1:2 molar ratio. Therefore, the number of moles of H2SO4 in the given sample can be calculated as:

moles H2SO4 = 2 × moles NaOH

moles NaOH = concentration of NaOH × volume of NaOH used = 0.150 M × 35.0 mL = 0.00525 moles

moles H2SO4 = 2 × 0.00525 = 0.0105 moles

The volume of H2SO4 used is 25.0 mL = 0.0250 L

Therefore, the concentration of H2SO4 is:

concentration of H2SO4 = moles H2SO4 / volume of H2SO4 used = 0.0105 moles / 0.0250 L = 0.420 M

So, the concentration of H2SO4 is 0.420 M.

Well, if the NaOH "completely neutralizes" the H2SO4, then we can assume they make a great pair. So, let's go on a little adventure to find the concentration of H2SO4!

We know that the NaOH solution is 0.150 M, which basically means it's got a lot of NaOH buddies in it. And since it took 35.0 mL of this NaOH solution to neutralize the H2SO4, we'll use that information too.

Now, let's convert those mLs into liters because we're feeling fancy like that. We'll divide 35.0 by 1000 to get 0.035 L.

To find the concentration of the H2SO4, we can use the equation: (concentration of NaOH) x (volume of NaOH) = (concentration of H2SO4) x (volume of H2SO4).

Since the volume of the NaOH is 0.035 L and the concentration of the NaOH is 0.150 M, we can plug these values into the equation.

0.150 M x 0.035 L = (concentration of H2SO4) x (25.0 mL converted to liters)

Simplifying the equation a bit, we have:

0.150 M x 0.035 L = (concentration of H2SO4) x 0.025 L

Now, let's solve for the concentration of H2SO4!

(concentration of H2SO4) = (0.150 M x 0.035 L) / 0.025 L

Calculating this out, we get:

(concentration of H2SO4) = 0.21 M

So, the concentration of H2SO4 is 0.21 M, making it one amusing and mildly acidic solution!

To find the concentration of the H2SO4, you need to first determine the amount of NaOH that reacts with the acid. This can be done using the information provided about the volume and concentration of NaOH used.

Let's start by setting up a balanced chemical equation for the reaction between H2SO4 and NaOH:

H2SO4 + 2NaOH -> Na2SO4 + 2H2O

From the equation, we can see that one mole of H2SO4 reacts with 2 moles of NaOH. Therefore, the moles of NaOH used can be determined by multiplying the volume of NaOH used (35.0 mL) by its concentration (0.150 M):

moles NaOH = volume NaOH (L) × concentration NaOH (mol/L)

Since the volume of NaOH is provided in milliliters, we need to convert it to liters by dividing by 1000:

moles NaOH = (35.0 mL ÷ 1000) L × 0.150 mol/L

Next, we can use the stoichiometry of the balanced equation to determine the moles of H2SO4 that reacted. Since the mole ratio of NaOH to H2SO4 is 2:1, the moles of H2SO4 can be calculated using the moles of NaOH:

moles H2SO4 = (moles NaOH ÷ 2)

Now we can calculate the concentration of H2SO4. Given that the volume of H2SO4 is 25.0 mL, we need to convert it to liters by dividing by 1000:

concentration H2SO4 (mol/L) = (moles H2SO4) ÷ (volume H2SO4 (L))

Substituting in the calculated values:

concentration H2SO4 (mol/L) = (moles H2SO4) ÷ (25.0 mL ÷ 1000)

Simplifying further:

concentration H2SO4 (mol/L) = (moles H2SO4) ÷ (0.025 L)

Now, you can substitute the value for moles H2SO4, which was calculated earlier:

concentration H2SO4 (mol/L) = (moles NaOH ÷ 2) ÷ (0.025 L)

Finally, substitute the value for moles NaOH:

concentration H2SO4 (mol/L) = [(35.0 mL ÷ 1000) L × 0.150 mol/L ÷ 2] ÷ (0.025 L)

Evaluating this expression will give you the concentration of H2SO4.

H2SO4 + 2NaOH ==> Na2SO4 + 2H2O

mols NaOH = M x L = ?
mols H2SO4 = 2x mols NaOH (look at the coefficients n the balanced equation.)_
M H2SO4 = mols H2SO4/L H2SO4