100 mL of LiNO₃(aq),0.241 M is mixed with 240 mL of 0.618 M Ca(NO₃)₂(aq).What is the final concentration of NO₃^-(aq)in the solution? A)0.508 M B)1.01 M C)0.943 M D)0.756 M E)1.38 M

Question

Quizplus · Accepted Answer

First, we need to balance the chemical equation for the reaction between LiNO₃(aq) and Ca(NO₃)₂(aq): 2 LiNO₃(aq) + 3 Ca(NO₃)₂(aq) → Ca3(NO₃)2(aq) + 6 Li^-(aq) Next, we can use the formula M1V1 = M2V2 to find the final concentration of NO₃^-(aq). We can choose to solve for M2, the final concentration: M2 = (M1V1 + M2V2) / (V1 + V2) M1 = 0.241 M (from the LiNO₃(aq) solution) V1 = 0.100 L (100 mL) M2 = ? (what we are solving for) V2 = 0.240 L (240 mL) For Ca(NO₃)₂(aq), we need to first find the moles present in the solution: 0.618 M = moles / 1 L moles = 0.618 mol Since the balanced equation shows a 2:3 ratio between LiNO₃(aq) and Ca(NO₃)₂(aq), we can use stoichiometry to find the moles of LiNO₃(aq) present in the solution: moles LiNO₃(aq) = (2/3) x moles Ca(NO₃)₂(aq) moles LiNO₃(aq) = (2/3) x 0.618 mol moles LiNO₃(aq) = 0.412 mol Now we can plug all the values into the M2 equation to solve for the final concentration of NO₃^-(aq): M2 = (0.241 M x 0.100 L + 0.412 mol / 0.340 L) = 0.943 M Therefore, the final concentration of NO₃^-(aq) in the solution is 0.943 M.

100 ML of LiNO3(aq),0

100 ML of LiNO₃(aq),0