Question

1.88gof potassium chloride is dissolved in300.mLof a23.0mMaqueous solution of silver nitrate.Calculate the final molarity of chloride anion in the solution. You can assume the volume of the solution doesn't change when the potassium chloride is dissolved in it.Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

The final molarity of chloride anion in the solution is 0 molar.

Explanation:

Mass of potassium chloride solution = 1.88 g

Moles of potassium chloride = $\frac{1.88 g}{74.5 g/mol}=0.02524 mol$

Moles of silver nitrate = n

Molarity of silver nitrate solution = M= 23.0 M

Volume of the silver nitrate solution ,V= 300.0 mL = 0.3 L

$moles(n)=Molarity(M)\times Volume (L)$

$n=23.0 M\times 0.3 L = 6.9 mol$

$AgNO_3(aq)+KCl(aq)\rightarrow AgCl(s)+KNO_3(aq)$

According to reaction 1 mol of potassium nitrate reacts with 1 mol of silver nitrate .

Then 0.02524 moles of potassium chloride will react with:

$\frac{1}{1}\times 0.02524 mol=0.02524 mol$ of silver nitrate.

As we can see that potassium chloride is in limiting amount due to which the potassium chloride will get completely converted into silver chloride and potassium nitrate.

Since , no potassium chloride will left after reaction which indicates that no chloride ions will be present after reaction.

$[Cl^-]=0 M$

The final molarity of chloride anion in the solution is 0 molar.