A volume of 70.0mL of aqueous potassium hydroxide(KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the solution if 19.2mL of 1.50M H2SO4 was needed? The equation is 2KOH+H2SO4->K2SO4+2H20

moles H2SO4 = M x L = ??

moles KOH = 2 x moles H2SO4.
M KOH = moles KOH/L KOH

vai se fuder ---- > vai tomar no cu

Well, well, well, looks like we've got ourselves a chemistry question! Don't worry, I won't turn into a clownfish while answering this one.

Alright, let's get down to business. We know that the balanced equation is 2KOH + H2SO4 -> K2SO4 + 2H2O. It's titration time!

We're given that 70.0 mL of KOH was titrated against 19.2 mL of 1.50 M H2SO4. So, using the balanced equation, we can see that the molar ratio between KOH and H2SO4 is 2:1.

This means that for every 2 moles of KOH, we need 1 mole of H2SO4. Since we have the volume and concentration of H2SO4, we can calculate the number of moles of H2SO4 used:

moles of H2SO4 = volume (in L) × concentration (in mol/L)
= 0.0192 L × 1.50 mol/L
= 0.0288 mol

Since the molar ratio is 1:2, we can determine the number of moles of KOH used:

moles of KOH = 2 × moles of H2SO4
= 2 × 0.0288 mol
= 0.0576 mol

Now let's find the molarity of the KOH solution:

molarity of KOH = moles of KOH / volume (in L)
= 0.0576 mol / 0.0700 L
= 0.823 M

So, the molarity of the KOH solution is approximately 0.823 M. And there you have it, chemistry made fun!

To find the molarity (M) of the sulfuric acid (H2SO4) solution, we can use the equation:

M1V1 = M2V2

where:
M1 is the molarity of the potassium hydroxide (KOH) solution
V1 is the volume of the KOH solution (70.0 mL)
M2 is the molarity of the sulfuric acid solution (unknown)
V2 is the volume of the sulfuric acid solution (19.2 mL)

Plugging in the given values, we have:

M1 = ?
V1 = 70.0 mL
M2 = 1.50 M
V2 = 19.2 mL

Rearranging the equation, we get:

M2 = (M1 * V1) / V2

Now we can substitute the known values into the equation and solve for M2:

M2 = (M1 * 70.0 mL) / 19.2 mL

Since we don't know the molarity of the potassium hydroxide solution (M1), we cannot directly calculate M2. However, we can use the balanced chemical equation to determine the ratio between the number of moles of KOH and H2SO4 in the reaction.

From the equation:
2 moles of KOH react with 1 mole of H2SO4

Let's say the molarity of the KOH solution is M1. The number of moles of KOH can be calculated using the formula:

moles of KOH = M1 * (volume of KOH solution in liters)

In this case, the volume of the KOH solution is 70.0 mL, which is equal to 0.0700 L (since 1 L = 1000 mL).

moles of KOH = M1 * 0.0700 L

Now, according to the balanced equation, we need half the number of moles of H2SO4 compared to KOH. Therefore:

moles of H2SO4 = (1/2) * moles of KOH

Now, we know that the moles of H2SO4 can be calculated from its molarity (M2) and volume (V2) using the formula:

moles of H2SO4 = M2 * (volume of H2SO4 solution in liters)

In this case, the volume of the H2SO4 solution is 19.2 mL, which is equal to 0.0192 L.

moles of H2SO4 = M2 * 0.0192 L

Since we have the relationship between the moles of KOH and H2SO4:

moles of KOH = 2 * moles of H2SO4

Using the above equations, we can set up an expression for M2:

M2 * 0.0192 L = 2 * M1 * 0.0700 L

Simplifying the equation:

M2 = (2 * M1 * 0.0700 L) / 0.0192 L

Now, we can substitute the given values of M1 and solve for M2:

M2 = (2 * 1.50 M * 0.0700 L) / 0.0192 L

M2 = 5.21 M

Therefore, the molarity of the sulfuric acid (H2SO4) solution is 5.21 M.

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