Final answer:
The concentration of nitrate ions after mixing 0.325 M Pb(NO3)2 with 0.225 M K3PO4 is calculated by doubling the moles of Pb(NO3)2 to account for the two nitrate ions per formula unit, and then dividing by the total volume. The closest available answer option is 0.326 M, although the precise calculation results in 0.378 M.
Step-by-step explanation:
The student is asking for the concentration of nitrate ions in the resulting solution after mixing 0.325 M Pb(NO3)2 with 0.225 M K3PO4.
To calculate the nitrate ion concentration, we must understand that each mole of Pb(NO3)2 contains two moles of nitrate ions. Therefore, when 355 ml of 0.325 M Pb(NO3)2 is mixed, the number of moles of nitrate ions is double the number of moles of Pb(NO3)2. First, we calculate the moles of nitrate ions in the Pb(NO3)2 solution:
Moles of nitrate ions in Pb(NO3)2 = 0.355 L * 0.325 M * 2 = 0.2305 moles
Then, we need to find the total final volume:
Total volume = 355 ml + 255 ml = 610 ml = 0.610 L
Now, the final concentration of nitrate ions will be:
Final concentration of nitrate ions = 0.2305 moles / 0.610 L = 0.378 M
However, option d. 0.326 M is the closest to the correct answer, indicating a possible miscalculation or misunderstanding in the question as provided. The final concentration of nitrate ions is not exactly represented in the options given to the student.