# Question deead

Aug 20, 2015

["OH"^(-)] = "0.0688 M"

#### Explanation:

FULL QUESTION

The ${K}_{s p}$ of ${\text{Ca(OH)}}_{2}$ is determined in the two following saturated solutions: water 100.0mL is solution no. (1) and saturated 50.0mL of 0.0500M NaOH is solution no. (2).

In each of the two saturated solutions, the concentration of ${\text{OH}}^{-}$ is determined by titration with standardised $\text{HCl}$ solution.

For saturated solution (2) it was found that 31.40 mL of 0.1096 M $\text{HCl}$ was required to neutralise the ${\text{OH}}^{-}$ ions. Calculate the concentration of ${\text{OH}}^{-}$ ions in the saturated solution.

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If I understood your problem correctly, you need to determine the concentration of the hydroxide ions in the second solution, the one that was titrated with hydrochloric acid.

The idea here is that a significant part of the number of moles of hydroxide ions present in this solution will come from the sodium hydroxide.

The molar solubility of calcium hydroxide in this solution will actually be lower than in the first solution because of the common ion effect.

So, what you need to do is determine how many moles of sodium hydroxide the second solution contained by using the number of moles of hydrochloric acid used in the titration.

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{\text{HCl" = "0.1906 M" * 31.40 * 10^(-3)"L" = "0.00344 moles HCl}}$

The net ionic equation for this neutralization reaction looks like this

$\text{H"^(+) + "OH"^(-) -> "H"_2"O}$

Since you have a $1 : 1$ mole ratio between the hydroxide ions and the added protons, you get that the solution contained

0.00344color(red)(cancel(color(black)("moles HCl"))) * "1 mole OH"^(-)/(1color(red)(cancel(color(black)("mole HCl")))) = "0.00344 moles OH"""^(-)

Since the second solution had a volume of 50.0 mL, the concentration of the hydroxide ions was

$\left[\text{OH"^(-)] = "0.00344 moles"/(50.0 * 10^(-3)"L") = color(green)("0.0688 M}\right)$

EXTRA STUFF

A more interesting problem would be to try and determine the ${K}_{s p}$ of calcium hydroxide using the second solution.

Assuming that you add some solid calcium hydroxide in 50.0 mL of sodium hydroxide solution, you would get that the initial concentration of the hydroxide ions was

["OH"^(-)]_0 = ["NaOH"] = "0.0500 M"

Since the final concentration of the hydroxide ions is known to be $\text{0.0688 M}$, you can use an ICE Table and the dissociation equilibrium of calcium hydroxide to figure out the value of ${K}_{s p}$

${\text{Ca(OH)"_2 " "rightleftharpoons" " "Ca"^(2+) " "+" " color(red)(2)"OH}}^{-}$

$\text{I}$ $\text{ " " } -$ $\text{ " " " " " " " " "0" " " " " " " " } 0.0500$

$\text{C" " " " } -$ $\text{ " " " " " " "(+x)" " " " " " } \left(+ \textcolor{red}{2} x\right)$

$\text{E" " " " } -$ $\text{ " " " " " " " " "x" " " " " " " } 0.0688$

This will allow you to determine the molar solubility of calcium hydroxide, which in this case is represented by $x$, in this solution.

$0.0688 = 0.0500 + 2 x$

$2 x = 0.0188 \implies x = \frac{0.0188}{2} = \text{0.0094 M}$

This means that calcium hydroxide's ${K}_{s p}$ will be

K_(sp) = x * 0.0688""^2

K_(sp) = 0.0094 * 0.0688""^2 = 4.4 * 10^(-5)#

This value is about ten times bigger than the accepted value of about $5.5 \cdot {10}^{- 6}$, which could mean that some of the collected values are incorrect.