Question

You could add HCl(aq) to the solution to precipitate out AgCl(s). What volume of a 0.200 M HCl(aq) solution is needed to precipitate the silver ions from 13.0 mL of a 0.180 M AgNO3 solution

Answer 1

Answer- Given, [HCl] = 0.200 M, [AgNO₃] = 0.180 M , volume = 13.0 mL

Reaction –

HCl(aq) + AgNO₃(aq) ------> AgCl(s) + HNO₃(aq)

Now we need to calculate moles of AgNO₃

Moles of AgNO₃ = 0.180 M * 0.0130 L

                             = 0.00234 moles

From the given balanced

1 moles of AgNO₃ = 1 moles of HCl

So , 0.00234 moles of AgNO₃ = ?

= 0.00234 moles of HCl

We are given molarity and we calculated moles of HCl

So, volume (L) = moles / molarity

                         = 0.00234 moles / 0.200 M

                         = 0.0117 L

So, 1 L = 1000 mL

0.0117 L = ?

= 11.7 mL

11.7 mL volume of a 0.200 M HCl(aq) solution is needed to precipitate the silver ions from 13.0 mL of a 0.180 M AgNO3 solution