Question

please help with equilibrium reaction and what happens to the OH ions in the solution

PART ASSOCIATION/DISSOCIATION OF A PRECIPITATE Data Collection 1. Get a clean test tube and place -1 mL (20-25 drops) of NaOH solution in the tube. 2. Slowly add drops of CaCl, solution until you notice a precipitate form in the tube. 3. Write a net ionic reaction for the formation of the precipitate below as an equilib- rium reaction: -NaOH ans + CaClzrag NaCl + Ca(OH) 4. 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? 5. Add -1 mL of water to a second test tube and add two drons of nhanalubia

Answer 1

3. First we need to write the balance reaction for NaOH and CaCl₂. The balanced reaction between NaOH and CaCl₂ will be

2 NaOH (aq) + CaCl₂ (aq) $\rightarrow$ 2 NaCl (aq) + Ca(OH)₂ (s)

To write the net ionic reaction for above reaction, first write reaction in ionic form.

2 Na⁺ (aq) + 2 OH^- (aq) + Ca²⁺ (aq) + 2 Cl^- (aq) $\rightarrow$ 2 Na⁺ (aq) + 2 Cl^- (aq) + Ca(OH)₂ (s)

Then cancel the spectator ion from above part. Since the spectator ions are Na⁺ and 2 Cl^- . After cancelling the spectator ion the reaction will look like

2 OH^- (aq) + Ca²⁺ (aq) $\rightarrow$ Ca(OH)₂ (s)

Therefore the net ionic reaction is

2 OH^- (aq) + Ca²⁺ (aq) $\rightarrow$ Ca(OH)₂ (s).

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Thanks.