Answer to Question #312011 in General Chemistry for Zellie

Question

You mix 50.0 mL of 0.165 M AgNO₃ with 25.0 mL of 0.302 M Ca(NO₃)₂. What is the concentration (M) of the NO₃^- anion in this solution mixture?

Answer 1

n(AgNO₃) = C(AgNO₃) * V(AgNO₃) = 0.165 * 0.05 = 0.00825 M;

n(NO₃^-)_AgNO3 = n(AgNO₃) = 0.00825 M;

n(Ca(NO₃)₂) = C(Ca(NO₃)₂) * V(Ca(NO₃)₂) = 0.302 * 0.025 = 0.00755 M;

n(NO₃^-)_Ca(NO3)2 = 2 * n(Ca(NO₃)₂) = 2 * 0.00755 = 0.0151 M;

n(NO₃^-)_total = n(NO₃^-)_AgNO3 + n(NO₃^-)_Ca(NO3)2 = 0.00825 + 0.0151 = 0.02335 M;

V_total = V(AgNO₃) + V(Ca(NO₃)₂) = 50.0 + 25.0 = 75.0 mL = 0.075 L;

C(NO₃^-)_total = n(NO₃^-)_total / V_total = 0.02335/0.075 = 0.3113 M.

Answer: 0.3113 M.