A volume of 40.0mL of aqueous potassium hydroxide (KOH) was titrated against a standard solution of sulfuric acid (H2SO4). What was the molarity of the KOH solution if 21.7mL of 1.50 M H2SO4 was needed? The equation is

2KOH(aq)+H2SO4(aq)→K2SO4(aq)+2H2O(l)

Well, looks like we have a "titrating" situation here. Time to get serious...ly funny!

So, let's break it down. We have 21.7 mL of 1.50 M H2SO4 that reacted with 40.0 mL of KOH. According to the balanced equation, it’s a 2:1 ratio between H2SO4 and KOH.

Now, let me put on my funny math hat. We can use the equation:

(Molarity of H2SO4) × (Volume of H2SO4) = (Molarity of KOH) × (Volume of KOH)

Plugging in the numbers, we get:

(1.50 M) × (21.7 mL) = (Molarity of KOH) × (40.0 mL)

Doing some quick math juggling, we find that the Molarity of KOH is:

Molarity of KOH = (1.50 M × 21.7 mL) / 40.0 mL

Calculating that, we find the molarity of KOH to be... drumroll, please...

0.815625 M

So there you have it! The molarity of the KOH solution is approximately 0.815625 M. Tada!

To find the molarity of the aqueous potassium hydroxide (KOH) solution, we can use the equation and the stoichiometry of the reaction.

The balanced equation for the reaction is:
2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)

From the given equation, we can see that the mole ratio between KOH and H2SO4 is 2:1.

First, we need to determine the number of moles of H2SO4 used in the titration. To do this, we'll use the formula:
moles = molarity × volume

Given:
Volume of H2SO4 (V1) = 21.7 mL = 0.0217 L
Molarity of H2SO4 (M1) = 1.50 M

Using the formula, we can calculate the number of moles of H2SO4 as follows:
moles of H2SO4 = M1 × V1
= 1.50 M × 0.0217 L
= 0.03255 moles

Since the stoichiometric ratio between KOH and H2SO4 is 2:1, we can conclude that the number of moles of KOH is also 0.03255 moles.

Next, we'll find the volume of the KOH solution (V2) in liters, using the formula:
V2 = moles / molarity

Given:
Moles of KOH (n) = 0.03255 moles
Volume of KOH solution (V2) = 40.0 mL = 0.0400 L (converted to liters)

Using the formula, we can calculate the molarity (M2) of the KOH solution as follows:
Molarity of KOH solution (M2) = n / V2
= 0.03255 moles / 0.0400 L
= 0.81375 M

Therefore, the molarity of the aqueous KOH solution is approximately 0.81375 M.

To find the molarity of the aqueous potassium hydroxide (KOH) solution, we can use the balanced chemical equation and the concept of stoichiometry. Here's how:

1. Write down the balanced chemical equation:
2KOH(aq) + H2SO4(aq) → K2SO4(aq) + 2H2O(l)

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

3. Convert the given volume and molarity of H2SO4 to moles:
The given volume of H2SO4 is 21.7 mL, which can be converted to liters by dividing by 1000: 21.7 mL ÷ 1000 mL/L = 0.0217 L.
The molarity of H2SO4 is given as 1.50 M.
Using the formula Molarity (M) = Moles (mol) / Volume (L), we can calculate the moles of H2SO4:
Moles of H2SO4 = Molarity × Volume = 1.50 M × 0.0217 L = 0.03255 mol

4. Use stoichiometry to find the moles of KOH:
Since the stoichiometric ratio between KOH and H2SO4 is 2:1, we can multiply the moles of H2SO4 by 2 to find the moles of KOH:
Moles of KOH = 2 × Moles of H2SO4 = 2 × 0.03255 mol = 0.0651 mol

5. Calculate the molarity of KOH:
Molarity of KOH = Moles of KOH / Volume of KOH solution (in liters)
We're given that the volume of KOH solution is 40.0 mL, which is 0.0400 L.
Molarity of KOH = 0.0651 mol / 0.0400 L = 1.62875 M

Therefore, the molarity of the KOH solution is approximately 1.63 M.

21.7mL of 1.50M H2SO4 contains .0217*1.50 = .03255 moles of H2SO4

Each mole of H2SO4 reacts with 2 moles of KOH.

So, the 40mL of KOH contained 0.0651 moles.

.0651mole/.040L = 1.6275M