Question

Calculate the concentration of an aqueous solution of Ca(OH)2Ca(OH)2 that has a pHpH of 12.11. Express your answer using two significant figures.

Answer 1

To find the concentration of Ca(OH)2 with a pH of 12.11, calculate the pOH, determine [OH-], and then halve it because Ca(OH)2 disassociates into two OH- ions. The concentration of Ca(OH)2 is 0.0065 M.

Step-by-step explanation:

To calculate the concentration of an aqueous solution of Ca(OH)2 with a pH of 12.11, we first need to calculate the pOH, which is the negative logarithm of the hydroxide ion concentration ([OH−]). Since pH + pOH = 14, we get:

pOH = 14 - pH = 14 - 12.11 = 1.89

Now we use the hydroxide ion concentration formula:

[OH−] = 10−pOH = 10−1.89 ≈ 1.29 x 10−2 M

Ca(OH)2 dissociates into one Ca2+ ion and two OH− ions. Therefore, the concentration of Ca(OH)2 is half the concentration of [OH−]:

[Ca(OH)2] = [OH−] / 2 = (1.29 x 10−2 M) / 2 ≈ 6.45 x 10−3 M

Thus, the concentration of Ca(OH)2 is 0.0065 M, rounded to two significant figures.

Answer 2

Answer: Concentration of
`Ca(OH)_2` is
`5.9* 10{-4}M`

Step-by-step explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

`pH=-\log [H^+]`

`pOH=-\log [OH^-]`

`12.11+pOH=14`

`pOH=1.89`

`1.89=-\log [OH^-]`

`[OH^-]=0.0129`

`Ca(OH)_2\rightarrow Ca^(2+)+2OH^(-)`

According to stoichiometry,

2 mole of
`OH^-` are produced by = 1 mole of
`Ca(OH)_2`

Thus 0.0129 moles of
`OH^-` are produced by =
`(1)/(2)* 0.0129=5.86* 10{-4}` moles of
`Ca(OH)_2`

The concentration of
`Ca(OH)_2` is
`5.9* 10{-4}M`