What volume of 0.0482 M Ba(OH)2 is required to neutralize exactly 16.10 mL of 0.141 M H3PO4? Phosphoric acid contains three acidic hydrogens.

Write the equation and balance it.

3Ba(OH)2 + 2H3PO4 ==> Ba3(PO4)2 + 6H2O

moles H3PO4 given = M x L = ??
Using the coefficients in the balancd equation, convert moles H3PO4 to moles Ba(OH)2.

Now M Ba(OH)2 = moles Ba(OH)2/L Ba(OH)2.
You have M and moles, solve for L.

Well, let me do some calculations in "Clown Bot" style:

Ba(OH)2 and H3PO4 are planning to go on a date, but they need to find the right volume of Ba(OH)2 to neutralize H3PO4. It's like finding the perfect amount of popcorn for a movie night!

First, let's find the moles of H3PO4. Since the concentration is given in Molarity (M), we can use the formula Molarity = moles/volume.

Moles of H3PO4 = 0.141 M x 0.01610 L
Moles of H3PO4 = 0.0022741 mol

Now, since H3PO4 contains three acidic hydrogens, it will require 3 moles of Ba(OH)2 to neutralize 1 mole of H3PO4.

To find the volume of Ba(OH)2, we can use the formula moles = Molarity x volume.

0.0022741 mol H3PO4 = (3 mol Ba(OH)2 / 1 mol H3PO4) x (0.0482 M Ba(OH)2) x (volume of Ba(OH)2)

Now, let's solve for the volume of Ba(OH)2:

volume of Ba(OH)2 = 0.0022741 mol H3PO4 / (3 mol Ba(OH)2 / 1 mol H3PO4 x 0.0482 M Ba(OH)2)
volume of Ba(OH)2 ≈ 0.0151 L

So, approximately 0.0151 liters (or 15.1 mL) of 0.0482 M Ba(OH)2 would be required to neutralize exactly 16.10 mL of 0.141 M H3PO4.

Remember, chemistry can be tricky, just like juggling. So, don't hesitate to consult your teacher or a real-life chemist to double-check!

To answer this question, we need to calculate the balanced chemical equation for the reaction between Ba(OH)2 and H3PO4 and then use stoichiometry to find the required volume of Ba(OH)2.

The balanced chemical equation for the reaction between Ba(OH)2 and H3PO4 is:

Ba(OH)2 + 2H3PO4 -> Ba(H2PO4)2 + 2H2O

From the balanced equation, we can see that 1 mole of Ba(OH)2 reacts with 2 moles of H3PO4.

First, let's calculate the number of moles of H3PO4:

Moles H3PO4 = Concentration (mol/L) * Volume (L)
= 0.141 M * (16.10 mL / 1000 mL/L)
= 0.0022731 mol

According to the balanced equation, 1 mole of Ba(OH)2 reacts with 2 moles of H3PO4. This means that we need half as many moles of Ba(OH)2 as there are moles of H3PO4.

Moles Ba(OH)2 = 0.0022731 mol / 2
= 0.00113655 mol

Finally, let's calculate the volume of Ba(OH)2 at a concentration of 0.0482 M:

Volume Ba(OH)2 = Moles / Concentration
= 0.00113655 mol / 0.0482 M
= 0.0236 L

However, the volume is given in liters, so let's convert it to milliliters:

Volume Ba(OH)2 = 0.0236 L * 1000 mL/L
= 23.6 mL

Therefore, approximately 23.6 mL of 0.0482 M Ba(OH)2 is required to neutralize exactly 16.10 mL of 0.141 M H3PO4.

To determine the volume of Ba(OH)2 needed to neutralize the given volume of H3PO4, you can use the stoichiometry of the balanced chemical equation and the given molarities of the two solutions.

First, let's write the balanced chemical equation for the reaction between Ba(OH)2 and H3PO4:

2 Ba(OH)2 + 3 H3PO4 → Ba3(PO4)2 + 6 H2O

From the balanced equation, we can see that it takes two moles of Ba(OH)2 to react with three moles of H3PO4.

Given:
Molarity of Ba(OH)2 solution (M1) = 0.0482 M
Volume of H3PO4 solution (V2) = 16.10 mL = 0.01610 L
Molarity of H3PO4 solution (M2) = 0.141 M

Now, we can calculate the volume of Ba(OH)2 solution needed using the following formula:

M1V1 = M2V2

Rearranging the formula, we get:

V1 = (M2 * V2) / M1

Substituting the values:

V1 = (0.141 M * 0.01610 L) / 0.0482 M
= 0.047 L

The volume of 0.0482 M Ba(OH)2 solution required to neutralize 16.10 mL of 0.141 M H3PO4 is 0.047 L (47 mL).