Problem: If 308 mL of 0.148 M NaCl solution is added to 228 mL of a 0.369 M NH4NO3 solution, the concentration of ammonium ions in the resulting mixture will be:A. 0.625 M B. 0.218 MC. 0.369 M D. 0.157 ME. 0 M

🤓 Based on our data, we think this question is relevant for Professor Taylor's class at UTD.

FREE Expert Solution

Calculate the NH4+ (ammonium ion) produced from 228 mL of 0.369 M NH4NOand 308 mL NaCl

Recall that molarity can be calculated as:

Molarity = moles of soluteL of solution


NH4NO3 dissociates in water as:

View Complete Written Solution
Problem Details

If 308 mL of 0.148 M NaCl solution is added to 228 mL of a 0.369 M NH4NO3 solution, the concentration of ammonium ions in the resulting mixture will be:

A. 0.625 M 

B. 0.218 M

C. 0.369 M 

D. 0.157 M

E. 0 M