Question

You have wells with different concentrations of NaOH (0.01M, 0.001 M, 0.0001 M, 0.00001M). You inject the wells with CO2 gas and observe the color changes (added universal color indicator prior).

Calcium carbonate is a very insoluble salt. If you added a few drops of 0.1M CaCl2 (aq) to the wells, what would you expect to see? How would the observations differ between the wells? Show chemical equations to support your predictions.

Answer 1

Different NaOH wells- As universal color indicator prior, the wells shows characteristics colors at their respective pHs

1) 0.01M

pOH=-log[OH-]=-log(0.01)=2

pH=14-2=12

color=dark purple

2)0.001 M

pOH=-log[OH-]=-log(0.001)=3

pH=14-3=11

color=purple

3)0.0001 M

pOH=-log[OH-]=-log(0.0001)=4

pH=14-4=10

color=navy blue

4)0.00001M

pOH=-log[OH-]=-log(0.00001)=5

pH=14-5=9

color=blue

When CO2 is added to the wells,the following reactions may occur

CO2(g) +H2OH2CO3(aq)

when CO2 is added ,the pH becomes low,(CO2 turns carbonic acid in aqueous medium) so the color changes according to the pH change.For pH=9 at 0.00001M well, the pH would shift towards greenish end.

2)Calcium carbonate if present in NaOH containing wells,and 0.1M CaCl2 further added.then the excess of Ca2+ ions decreases the solubility of CaCO3, due to common ion effect.

CaCO3Ca2+ +CO3^2-

CaCl2Ca2+ +2Cl^- (highly soluble in water so complete ionization occurs)

The equilibrium shifts towards the left, and more CaCO3 precipitates.