14.0 mL , of 0.53 M barium nitrate solution and 16.0 mL of 0.44 M sodium sulfate solution are mixed . Which ion is LEA ST abundant in solution? (A) Ba^2+ (B) NO_(3)^- (C) Na^+ (D) SO_(4)^2-

To determine which ion is least abundant in the solution, we need to calculate the moles of each ion present after mixing the solutions. The total volume of the mixed solution is the sum of the volumes of the two solutions.<br /><br />Total volume = 14.0 mL + 16.0 mL = 30.0 mL<br /><br />Next, we calculate the moles of each ion:<br /><br />1. Barium nitrate (Ba(NO₃)₂):<br /> - Molarity (M) = 0.53 M<br /> - Volume (V) = 14.0 mL = 0.014 L<br /> - Moles of Ba²⁺ = M × V = 0.53 × 0.014 = 0.00742 moles<br /><br />2. Sodium sulfate (Na₂SO₄):<br /> - Molarity (M) = 0.44 M<br /> - Volume (V) = 16.0 mL = 0.016 L<br /> - Moles of Na⁺ = 2 × (M × V) = 2 × (0.44 × 0.016) = 0.01392 moles<br /> - Moles of SO₄²⁻ = 1 × (M × V) = 0.44 × 0.016 = 0.00704 moles<br /><br />Now, we sum up the moles of each ion in the mixed solution:<br /><br />- Total moles of Ba²⁺ = 0.00742 moles<br />- Total moles of NO₃⁻ = 2 × 0.00742 = 0.01484 moles (since each Ba(NO₃)₂ dissociates into one Ba²⁺ and two NO₃⁻)<br />- Total moles of Na⁺ = 0.01392 moles<br />- Total moles of SO₄²⁻ = 0.00704 moles<br /><br />Comparing the moles of each ion, we find that the ion with the least number of moles is:<br /><br />(D) \( SO_{4}^{2-} \)<br /><br />Therefore, the ion that is least abundant in the solution is \( SO_{4}^{2-} \).