Get a clean test tube and place ~1 mL (20-25 drops) of NaOH solution in the tube

Question

Get a clean test tube and place ~1 mL (20-25 drops) of NaOH solution in the tube. Slowly add drops of CaCl_2 solution until you notice a precipitate form in the tube. Write a net ionic reaction for the formation of the precipitate below as an equilibrium reaction: What happened to the OH^- ions in solution? Are they all part of a precipitate? Or are some still in solution? What could you do to find out? Add ~ 1 mL of water to a second test tube and add two drops of plenolphtlulein indicator. What happened? Now add 3 drops of NaOH solution to the second test tube. Record your observations. Phenolphthalein is an indicator that turns pink in the presence OH^- ions. How do you interpret your observations?

Explanation / Answer

Ans. 3 NaOH(aq)           -------------> Na+(aq)    + OH-(aq)         - reaction 1

        CaCl2(aq) -------------> Ca2+(aq)      + 2 Cl-(aq)         - reaction 2

Now,

2 x reaction 1 and reaction 2 gives overall ionic reaction as follows-

            2 Na+(aq) + 2 OH-(aq) + Ca2+(aq) + 2 Cl-(aq) ------> 2 Na+(aq) + 2 Cl-(aq) + Ca(OH)2(s)   

Net ionic reaction (cancel the species on either side of the reaction)-

            2 OH-(aq) + Ca2+(aq) + ------> Ca(OH)2(s)   --- Net ionic reaction.

Ans. 4. 2 mol OH- ions react with 1 mol Ca2+ ions to form Ca(OH)2 which is precipitated as white mass in the tube.

Since, the solution is still pink (top pink), it indicates the presence of some OH- ions in the solution because phenolphthalein indicator is pink in alkaline medium (excess of OH- ions) only.

Ans. 5. No color change is observed. Water being neutral does not affect the pH. Thus, in absence of excess OH-, phenolphthalein indicator remains colorless in the solution.

Ans. 6. The solution turn pink. NaOH (a strong base) donates OH- ions in aqueous solution making it alkaline. Thus, phenolphthalein indicator turns pink in alkaline solution- causing the solution to appear pink, too.

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