Question

8. Consider what happens when Ca(OH)2, dissolves in water. Write the balanced equation for the reaction. What can we say about the size of the equilibrium constant for this reaction? What is the [OH-] in a 3.25 mM solution of calcium hydroxide? POH? d. What is the pH of the above solution? [H30]?! What is the concentration of a calcium hydroxide solution with pH = 9.25?

Answer 1

1) Ca(OH)₂ is a strong base; that means it totally dissociates in water, in the following way:

$Ca(OH)_{2}\leftrightharpoons Ca^{2+}+2OH^{-}$

Where the equilibrium equilibrium is extremely displaced towards the products and the reaction is considered to occur fully. That is: Ca(OH)₂ dissociation can be considered complete.

2) This last means that the equilibrium constant is enormous in this case. It can actually be considered as infinitely large.

3) To calculate the concentration of OH^-, we must bear in mind that each mole of Ca(OH)₂ that is dissolved results in 2 moles of hydroxide ions. Which means that their concentration will be double the original concentration of Ca(OH)₂. In this case:

$[OH^{-}]=2\cdot [Ca(OH)_{2}]=2\cdot 3.25mM=6.5mM=0.0065M$

pOH is calculated as:

$pOH=-log[OH^{-}]=-log(0.0065)=2.19$

4) pH can be calculated directly from pOH by using the relation:

$pOH+pH=14$

So:

$pH=14-pOH=14-2.19=11.81$

5) This is the reverse of what we just did. If pH is 9.25, then pOH is:

$pOH=14-pH=14-9.25=4.75$

And the concentration of OH^- is given by:

$[OH^{-}]=10^{-pOH}=10^{-4.75}=1.78x10^{-5}M$

And, as we saw, the concentration of Ca(OH)₂ is half the concentration of OH^-, so:

$[Ca(OH)_{2}]=\frac{[OH^{-}]}{2}=\frac{1.78x10^{-5}}{2}=8.89x10^{-6}M$