We need to first find the molarity of Ba(OH₂) solution.
A mass of 3.24 mg is dissolved in 1 L solution.
Ba(OH)₂ moles dissolved - 3.24 x 10⁻³ g/171.3 g/mol = 1.90 x 10⁻⁵ mol
dissociaton of Ba(OH)₂ is as follows;
Ba(OH)₂ --> Ba²⁺ + 2OH⁻
1 mol of Ba(OH)₂ dissociates to form 2OH⁻ ions.
Therefore [OH⁻] = (1.90 x 10⁻⁵)x2 = 3.8 x 10⁻⁵ M
pOH = -log[OH⁻]
pOH = -log (3.8 x 10⁻⁵)
pOH = 4.42
pH + pOH = 14
therefore pH = 14 - 4.42
pH = 9.58
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Answer:
At the cathode during the electrolysis of an aqueous solution of magnesium iodide, MgI2 , 2I−(aq) is produced
Explanation:
At cathode, reduction reaction takes place.
The dissociation of MgI2 in aqueous solution is Mg2+(aq) and 2I−(aq)
Here, the Iodine reduces to 2I−(aq) from state of 0 (MgI2) to state of -1 (2I−(aq))
Hence, at the cathode during the electrolysis of an aqueous solution of magnesium iodide, MgI2 , 2I−(aq) is produced
