Question

question 5 (5 points)
what is the total concentration of ions in a 0.10 m solution of barium chloride?
○ a) 0.10 m
○ b) 0.20 m
○ c) 0.30 m
○ d) 0.40 m

Explanation:

Step1: Write the dissociation equation

Barium chloride (\(BaCl_2\)) dissociates in water as: \(BaCl_2
ightarrow Ba^{2+} + 2Cl^-\)

Step2: Determine ion concentrations

From the equation, 1 mole of \(BaCl_2\) gives 1 mole of \(Ba^{2+}\) and 2 moles of \(Cl^-\). For a \(0.10\ M\) \(BaCl_2\) solution:

• \([Ba^{2+}] = 0.10\ M\)
• \([Cl^-] = 2\times0.10\ M = 0.20\ M\)

Step3: Calculate total ion concentration

Total ion concentration = \([Ba^{2+}] + [Cl^-] = 0.10\ M + 0.20\ M = 0.30\ M\) (Wait, original selected option was wrong. Let's recalculate.) Wait, no: Wait, \(BaCl_2\) dissociates into 1 \(Ba^{2+}\) and 2 \(Cl^-\), so total moles of ions per mole of \(BaCl_2\) is \(1 + 2 = 3\)? No, wait, concentration: 1 mole \(BaCl_2\) gives 1 mole \(Ba^{2+}\) (concentration \(0.10\ M\)) and 2 moles \(Cl^-\) (concentration \(2\times0.10 = 0.20\ M\)). So total ion concentration is \(0.10 + 0.20 = 0.30\ M\). So the correct answer should be c) 0.30 M.

Answer:

c) 0.30 M