barium hydroxide is an ionic compound built from a metal cation (barium) and the polyatomic hydroxide ion. Its formula, charge balance, and aqueous behavior follow standard rules for salts with polyatomic ions.
Key results: \(\mathrm{Ba(OH)_2}\); ions \(\mathrm{Ba^{2+}}\) and \(\mathrm{OH^-}\); dissociation \(\mathrm{Ba(OH)_2 \rightarrow Ba^{2+} + 2\,OH^-}\); molar mass \(\approx 171.35\ \text{g/mol}\).
Step 1: Identify the ions and their charges
Barium is a Group 2 metal, so it forms a \(+2\) cation:
\[ \mathrm{Ba \rightarrow Ba^{2+}} \]
Hydroxide is the polyatomic ion:
\[ \mathrm{OH^-} \]
Step 2: Build a neutral formula from charge balance
A neutral ionic compound has total charge \(0\). One \(\mathrm{Ba^{2+}}\) requires two \(\mathrm{OH^-}\) ions to balance charge:
\[ (+2) + 2\cdot(-1) = 0 \]
Therefore, the correct chemical formula is:
\[ \mathrm{Ba(OH)_2} \]
Parentheses indicate that the entire \(\mathrm{OH}\) group occurs twice.
Step 3: Write the dissociation in water (stoichiometry of ions)
In aqueous solution, barium hydroxide separates into one barium(II) ion and two hydroxide ions:
\[ \mathrm{Ba(OH)_2(s) \rightarrow Ba^{2+}(aq) + 2\,OH^-(aq)} \]
Step 4: Calculate the molar mass of \(\mathrm{Ba(OH)_2}\)
Add the atomic masses of 1 barium, 2 oxygen, and 2 hydrogen atoms. Using common periodic-table values \(M(\mathrm{Ba}) = 137.33\ \text{g/mol}\), \(M(\mathrm{O}) = 16.00\ \text{g/mol}\), \(M(\mathrm{H}) = 1.008\ \text{g/mol}\):
\[ M(\mathrm{Ba(OH)_2}) = 137.33 + 2\cdot(16.00 + 1.008) = 137.33 + 2\cdot 17.008 = 137.33 + 34.016 = 171.346\ \text{g/mol} \]
Rounded to typical precision: \(M(\mathrm{Ba(OH)_2}) \approx 171.35\ \text{g/mol}\).
| Component | Count | Atomic mass (g/mol) | Contribution (g/mol) |
|---|---|---|---|
| Barium (Ba) | \(1\) | \(137.33\) | \(1 \cdot 137.33 = 137.33\) |
| Oxygen (O) | \(2\) | \(16.00\) | \(2 \cdot 16.00 = 32.00\) |
| Hydrogen (H) | \(2\) | \(1.008\) | \(2 \cdot 1.008 = 2.016\) |
| Total | \(171.346\ \text{g/mol} \approx 171.35\ \text{g/mol}\) |
Step 5: A short stoichiometry application (hydroxide produced)
Because \(\mathrm{Ba(OH)_2 \rightarrow Ba^{2+} + 2\,OH^-}\), each mole of barium hydroxide produces 2 moles of hydroxide ions:
\[ n(\mathrm{OH^-}) = 2 \cdot n(\mathrm{Ba(OH)_2}) \]
Example: if \(n(\mathrm{Ba(OH)_2}) = 0.250\ \text{mol}\), then
\[ n(\mathrm{OH^-}) = 2 \cdot 0.250 = 0.500\ \text{mol} \]
Naming note
The name barium hydroxide follows ionic naming rules: the metal name “barium” + the polyatomic ion name “hydroxide.” Barium has a fixed \(+2\) charge in typical ionic compounds, so a Roman numeral is not required in the name.