158 mL of a 0.148M NaCl solution is added to 228 mL of a 0.369M NH₄NO₃ solution. The concentration of ammonium ions in the resulting mixture is A) 0.157 M. B) 0.218 M. C) 0.625 M. D) 0.369 M. E) 0 M.

Question

Quizplus · Accepted Answer

First, we need to write out the balanced chemical equation for the reaction between NaCl and NH₄NO₃: NaCl + NH4NO3 → NaNO3 + NH4Cl Next, we can use the formula: M1V1 + M2V2 = M3V3 where M1 and V1 are the molarity and volume of the NaCl solution, M2 and V2 are the molarity and volume of the NH₄NO₃ solution, and M3 is the concentration we are looking for (the concentration of ammonium ions) and V3 is the total volume of the mixture. Plugging in the values, we get: (0.148 M)(0.158 L) + (0.369 M)(0.228 L) = M3(0.158 L + 0.228 L) 0.050584 + 0.084012 = M3(0.386 L) 0.134596 = M3(0.386 L) M3 = 0.3489 M However, we need to remember that the balanced chemical equation shows a 1:1 ratio of ammonium ions to NaCl. Therefore, the concentration of ammonium ions in the mixture is equal to the concentration of NaCl, which is 0.148 M. Therefore, the answer is B) 0.218 M is incorrect as it is not equal to 0.148 M.

158 ML of a 0