Question

What is the silver ion concentration in a solution prepared by mixing 425 mL 0.397 M silver nitrate with 427 mL 0.459 M sodium phosphate

Answer 1

Answer:

0 M is the silver ion concentration in a solution prepared mixing both the solutions.

Explanation:

$molarity=\frac{\text{Moles of solute}}{\text{Volume of solution (L)}}$

Moles of silver nitrate = n

Volume of the solution = 425 mL = 0.425  L (1 mL = 0.001 L)

Molarity of the silver nitrate solution = 0.397 M

$n=0.397 M\times 0.425 L=0.1687 mol$

Moles of sodium phosphate = n'

Volume of the sodium phosphate solution = 427 mL = 0.427  L (1 mL = 0.001 L)

Molarity of the sodium phosphate solution = 0.459 M

$n'=0.459 M\times 0.427 L=0.1960 mol$

$3AgNO_3+Na_3PO_4\rightarrow Ag_3PO_4+3NaNO_3$

According to reaction, 3 moles of silver nitrate reacts with 1 mole of sodium phosphate, then 0.1687 moles of silver nitrate will recat with :

$\frac{1}{3}\times 0.1687 mol=0.05623 mol$ of sodium phosphate

This means that only 0.05623 moles of sodium phosphate will react with all the 0.1687 moles of silver nitrate , making silver nitrate limiting reagent and sodium phosphate as an excessive reagent.

So, zero moles of silver nitrate will be left in the solution after mixing of the both solutions and hence zero moles of silver ions will left in the resulting solution.

0 M is the silver ion concentration in a solution prepared mixing both the solutions.