A sample of solid Ca(OH)2 is stirred in water at 30°C until the solution contains as much dissolved Ca(OH)2 as it can hold. A 100.-mL sample of this solution is withdrawn and titrated with 5.00 X 10^-2 M HBr. It requires 48.8 mL of the acid solution for neutralization. What is the molarity of the Ca(OH)2 solution?

Balanced equation:

2HBr + Ca(OH)2 → CaBr2 + 2H2O
1mol HBr reacts with 2mol Ca(OH)2
Use equation:
M1V1 = M2V2 *2
0.05 * 48.8 = M2*100*2
M2 = 0.05*48.8/(100*2)
M2 = 0.0122M Ca(OH)2 solution. answer to first question.

Molar mass Ca(OH)2 = 74.0932 g/mol
0.0122M solution contains: 74.0932*0.0122 *100/1000 = 0.09g/100ml

Solubility of Ca(OH)2 at 30°C = 0.09g / 100ml solution.

1. Write the balanced equation for the titration.

2. Calculate moles HBr used in the titration. moles = M x L.
3. Using the coefficients in the balanced equation from step 1, convert moles HBr to moles Ca(OH)2.
4. You want molarity Ca(OH)2. By definition, M = moles/L; therefore, use moles from step 3 and divide by 0.1 L.

Well, that's quite a specific question! It seems like you're dealing with some fun chemistry calculations. But here's a little joke for you before we dive in:

Why do chemists like nitrates so much?

Because they're cheaper than day rates!

Now, back to your question. To find the molarity of the Ca(OH)2 solution, we need to use the titration data. We can start by writing a balanced chemical equation for the reaction:

Ca(OH)2 + 2HBr -> CaBr2 + 2H2O

According to the equation, 1 mole of Ca(OH)2 reacts with 2 moles of HBr. So, using stoichiometry, we can calculate the number of moles of HBr used in the titration:

48.8 mL HBr * (5.00 x 10^-2 mol/L) = 2.44 x 10^-3 moles of HBr

Since 1 mole of Ca(OH)2 reacts with 2 moles of HBr, we know that 2.44 x 10^-3 moles of HBr reacted with 1/2 x 2.44 x 10^-3 moles of Ca(OH)2.

Now, let's find the volume of the Ca(OH)2 solution that was used in the titration:

100 mL of Ca(OH)2 solution * (1 L / 1000 mL) = 0.1 L

With this information, we can calculate the molarity of the Ca(OH)2 solution:

Molarity = moles / volume

Molarity = (1/2 * 2.44 x 10^-3) moles / 0.1 L

Molarity ≈ 0.0244 M

So, the molarity of the Ca(OH)2 solution is approximately 0.0244 M. I hope this helps, and remember, chemistry is all about mixing things up in the lab and having fun with it!

To find the molarity (M) of the Ca(OH)2 solution, we can use the formula:

M1V1 = M2V2

where M1 is the molarity of the HBr solution, V1 is the volume of HBr solution used for neutralization, M2 is the molarity of the Ca(OH)2 solution, and V2 is the volume of the Ca(OH)2 solution used in the titration.

Given:
M1 = 5.00 x 10^-2 M
V1 = 48.8 mL = 48.8 x 10^-3 L (convert mL to L)
V2 = 100 mL = 100 x 10^-3 L (convert mL to L)

Now, substitute the values into the formula:

(5.00 x 10^-2 M)(48.8 x 10^-3 L) = M2(100 x 10^-3 L)

Simplify the equation and solve for M2:

M2 = (5.00 x 10^-2 M)(48.8 x 10^-3 L) / (100 x 10^-3 L)

M2 = 2.44 x 10^-3 mol / 0.1 L

M2 = 0.0244 M

Therefore, the molarity of the Ca(OH)2 solution is 0.0244 M.

To determine the molarity of the Ca(OH)2 solution, you can use the equation:

Molarity (M1) x Volume (V1) = Molarity (M2) x Volume (V2)

Here, M1 is the unknown molarity of the Ca(OH)2 solution, V1 is the volume of the Ca(OH)2 solution withdrawn (100 mL), M2 is the molarity of the HBr solution (5.00 x 10^-2 M), and V2 is the volume of the HBr solution required for neutralization (48.8 mL).

Rearranging the equation to solve for M1:

M1 = (M2 x V2) / V1

Substituting the given values:

M1 = (5.00 x 10^-2 M) x (48.8 mL) / (100 mL)

Note that we need to convert mL to L since molarity is expressed in moles per liter:

M1 = (5.00 x 10^-2 M) x (48.8 mL ÷ 1000 mL/L) / (100 mL ÷ 1000 mL/L)

Now, let's calculate the value:

M1 = (5.00 x 10^-2 M) x (48.8 x 10^-3 L) / (100 x 10^-3 L)

M1 = 2.44 x 10^-3 M

Therefore, the molarity of the Ca(OH)2 solution is 2.44 x 10^-3 M.