Question
If BaCl2(aq) + NaOH (aq) react , do precipitate formed? If sure then write the balanced net ionic equation for the precipitate formed.
Answers
GPT-4o mini
When barium chloride (BaCl₂) reacts with sodium hydroxide (NaOH), a double displacement reaction occurs. In this reaction, barium hydroxide (Ba(OH)₂) is formed, which is a sparingly soluble compound that precipitates out of solution. The sodium chloride (NaCl) remains dissolved in the aqueous solution.
The balanced molecular equation for the reaction is:
\[ \text{BaCl}_2 (aq) + 2 \text{NaOH} (aq) \rightarrow \text{Ba(OH)}_2 (s) + 2 \text{NaCl} (aq) \]
To write the net ionic equation, we first break down the soluble strong electrolytes into their respective ions:
- \( \text{BaCl}_2 (aq) \) dissociates into \( \text{Ba}^{2+} (aq) \) and \( 2 \text{Cl}^- (aq) \)
- \( \text{NaOH} (aq) \) dissociates into \( \text{Na}^+ (aq) \) and \( \text{OH}^- (aq) \)
- \( \text{NaCl} (aq) \) dissociates into \( \text{Na}^+ (aq) \) and \( \text{Cl}^- (aq) \)
Now, writing the complete ionic equation:
\[ \text{Ba}^{2+} (aq) + 2 \text{Cl}^- (aq) + 2 \text{Na}^+ (aq) + 2 \text{OH}^- (aq) \rightarrow \text{Ba(OH)}_2 (s) + 2 \text{Na}^+ (aq) + 2 \text{Cl}^- (aq) \]
Now we can cancel the spectator ions (\(2 \text{Na}^+\) and \(2 \text{Cl}^-\)) that appear on both sides of the equation:
The net ionic equation is:
\[ \text{Ba}^{2+} (aq) + 2 \text{OH}^- (aq) \rightarrow \text{Ba(OH)}_2 (s) \]
Thus, the precipitate formed is barium hydroxide, and the net ionic equation shows that barium ions combine with hydroxide ions to form the insoluble precipitate.
The balanced molecular equation for the reaction is:
\[ \text{BaCl}_2 (aq) + 2 \text{NaOH} (aq) \rightarrow \text{Ba(OH)}_2 (s) + 2 \text{NaCl} (aq) \]
To write the net ionic equation, we first break down the soluble strong electrolytes into their respective ions:
- \( \text{BaCl}_2 (aq) \) dissociates into \( \text{Ba}^{2+} (aq) \) and \( 2 \text{Cl}^- (aq) \)
- \( \text{NaOH} (aq) \) dissociates into \( \text{Na}^+ (aq) \) and \( \text{OH}^- (aq) \)
- \( \text{NaCl} (aq) \) dissociates into \( \text{Na}^+ (aq) \) and \( \text{Cl}^- (aq) \)
Now, writing the complete ionic equation:
\[ \text{Ba}^{2+} (aq) + 2 \text{Cl}^- (aq) + 2 \text{Na}^+ (aq) + 2 \text{OH}^- (aq) \rightarrow \text{Ba(OH)}_2 (s) + 2 \text{Na}^+ (aq) + 2 \text{Cl}^- (aq) \]
Now we can cancel the spectator ions (\(2 \text{Na}^+\) and \(2 \text{Cl}^-\)) that appear on both sides of the equation:
The net ionic equation is:
\[ \text{Ba}^{2+} (aq) + 2 \text{OH}^- (aq) \rightarrow \text{Ba(OH)}_2 (s) \]
Thus, the precipitate formed is barium hydroxide, and the net ionic equation shows that barium ions combine with hydroxide ions to form the insoluble precipitate.