Question

In this experiment we will be using a 0.05 M solution of HCl to determine the concentration of hydroxide (OH-) in a saturated solution of calcium hydroxide. The initial volume of "filtered" calcium hydroxide solution used in this example will be 36.0 mL. If it takes 16 mL of the HCl solution to reach the equivalence point of the titration, how many moles of OH- were present in this sample of calcium hydroxide?

Answer 1

Answer: The moles of hydroxide ions present in the sample is 0.0008 moles

Explanation:

To calculate the concentration of acid, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is HCl.

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.05M\\V_1=16mL\\n_2=2\\M_2=?M\\V_2=36.0mL$

Putting values in above equation, we get:

$1\times 0.05\times 16=2\times M_2\times 36\\\\M_2=\frac{1\times 0.05\times 16}{2\times 36}=0.011M$

To calculate the moles of hydroxide ions, we use the equation used to calculate the molarity of solution:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

Molarity of solution = 0.011 M

Volume of solution = 36.0 mL

Putting values in above equation, we get:

$0.011=\frac{\text{Moles of }Ca(OH)_2\times 1000}{36}\\\\\text{Moles of }Ca(OH)_2=\frac{0.011\times 36}{1000}=0.0004mol$

1 mole of calcium hydroxide produces 1 mole of calcium ions and 2 moles of hydroxide ions.

Moles of hydroxide ions = (0.0004 × 2) = 0.0008 moles

Hence, the moles of hydroxide ions present in the sample is 0.0008 moles