A solution is 0.660 M in HNO₃(aq)and 1.02 M in Ca(NO₃)₂(aq).What is the molarity of NO₃^-(aq)in the solution? A)2.70 M B)1.68 M C)1.02 M D)0.660 M E)0.360 M

Question

Quizplus · Accepted Answer

The molarity of $NO_3^-$ in the solution can be calculated by considering the contributions from both $HNO_3$ and $Ca(NO_3)_2$. $HNO_3$ contributes 0.660 M of $NO_3^-$ directly. $Ca(NO_3)_2$ dissociates into one $Ca^{2+}$ and two $NO_3^-$ ions, contributing $2 \times 1.02$ M = 2.04 M of $NO_3^-$. Adding these contributions together gives $0.660 \, M + 2.04 \, M = 2.70 \, M$ of $NO_3^-$ in the solution.

A Solution Is 0