Question

A 20.00 ml sample of calcium hydroxide is titrated with 0.0100 M HCl until the color of the phenolphthalein indicator just disappears. The following readings were made on the buret: Initial reading = 2.77 ml Final reading = 30.23 ml What is the calculated Ksp of the calcium hydroxide?

Answer 1

`1.29x10^-^6`

Step-by-step explanation:

We have to start with the ionization reaction of
`Ca(OH)2` so:

`Ca(OH)_2~-->~Ca^+^2~+~2OH^-^1` Reaction 1

With this equation we can write the Ksp expression:

`Ksp~=~[Ca^+^2][OH^-^1]^2`

So, if we want to calculate the Ksp of we have to know the
`OH^-^1` and
`Ca^+^2` concentrations. These concentrations can be calculate using the titration procedure. So, we have to start with the ionic reaction between HCl and Ca(OH)2, so:

`HCl~->~H^+~+~Cl^-` Reaction 2

`Ca(OH)_2~->~2OH^-~+~Ca^+^2` Reaction 3

`H^+~+~OH^---->~H_2O` Reaction4

The volume of HCl used in the experiment is the substraction between the inital and final reading:

`30.23~mL~-~2.77~mL~=~27.46~mL`

With this volume (27.46 mL = 0.02746 L) we can calculate the moles with the molarity equation, so:

`M=(mol)/(L)`

`0.01~M=(mol)/(0.02746~L)`

`mol~=~0.01*0.02746~=~0.0002746~mol~of~HCl`

The molar ratio between H+ and OH- is 1:1, therefore:

`0.0002746~mol~of~HCl=0.0002746~mol~of~OH^-`

Finally we have to calculate the concentration of OH- using the volume (20 mL= 0.02 L) and moles of OH-

`M=(0.0002746~mol~of~OH^-)/(0.02~L)`

`M=~0.01373~M~of~OH^-`

We already have the concentration of OH-the next step is the calculation of the concentration of
`Ca^+^2`. To do this, we have to use the molar ratio between
`Ca^+^2` and
`OH^-` (reaction 1), the molar ratio is 1:2 therefore:

`0.01373~M~of~OH^-(1)/(2)=~0.006865~M~of~Ca^+^2`

With these values now we can calculate the Ksp:

`Ksp=~[0.006865][0.01373]^2~=~1.29x10^-^6`

The Ksp value for the
`Ca(OH)_2` is
`1.29x10^-^6`