Question

A 25.0 mL sample of a saturated C a ( O H ) 2 solution is titrated with 0.023 M H C l , and the endpoint is reached after 38.0 mL of titrant are dispensed. Based on this data, what is the concentration (M) of C a ( O H ) 2

Answer 1

Answer:

The concentration of $Ca(OH)_2$ solution is 0.01748M .

Explanation:

Let us calculate -

Firstly , lets calculate $H^+ and OH^- ions$

Given - Volume of sample of $Ca(OH)_2 solution$ $(V_1)$ = 25.0ml

             Volume of sample of HCL solution $(V_2)$ = 38.0ml

             Concentration of HCL solution $(C_2)$ = 0.023M

            Concentration of Ca(OH) solution $(C_1)$ = ?

Now,

$C_1\times V_1 = C_2\times V_2$

Putting the given values-

$C_1\times25= 0.023\times38.0$

$C_1=0.03496M$

Thus, the concentration of $OH^-$ ions is 0.03496M

For, concentration of $Ca(OH)_2=0.5\times[OH]^-$

                                                 = 0.5 x 0.03496M

                                            = 0.01748M

Thus , concentration of $Ca(OH)_2$ solution is 0.01748M.

Hence , the answer is 0.01748M

           

Answer 2

The concentration of the Ca(OH)₂ solution is 0.0175 M

From the question,

We are to determine the concentration of the Ca(OH)₂ solution

First, we will write a balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

2HCl + Ca(OH)₂ → CaCl₂ + 2H₂O

This means,

2 moles of HCl is required to completely neutralize 1 mole of Ca(OH)₂

Using the formula

$\frac{C_{A}V_{A} }{C_{B}V_{B}} = \frac{n_{A}}{n_{B}}$

Where $C_{A}$ is the concentration of acid

$V_{A}$ is the volume of acid

$C_{B}$ is the concentration of base

$V_{B}$ is the volume of base

$n_{A}$ is the mole ratio of acid

$n_{B}$ is the mole ratio of base

From the question,

$C_{A} = 0.023 \ M$

$V_{A} = 38.0 \ mL$

$V_{B} = 25.0 \ mL$

From the balanced chemical equation

$n_{A} = 2$

$n_{B} = 1$

Putting the above parameters into the formula, we get

$\frac{0.023 \times 38.0}{C_{B} \times 25.0}= \frac{2}{1}$

Then, we get

$C_{B} \times 50.0 = 0.023 \times 38.0$

∴ $C_{B} = \frac{0.023 \times 38.0}{50.0}$

$C_{B} = \frac{0.874}{50.0}$

$C_{B} = 0.01748 \ M$

$C_{B} \approx 0.0175 \ M$

Hence, the concentration of the Ca(OH)₂ solution is 0.0175 M

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