Question
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 A 100.-mL sample of this solution is withdrawn and titrated with 5.00 X 10^-2 M HBr.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?
Answers
DrBob222
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.
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.
Shajed
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.
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.