Question

How much of a 1.30 M sodium chloride solution in milliliters is required to completely precipitate all of the silver in 15.0 mL of a 0.40 M silver nitrate solution?

Answer 1

Number of moles of silvernitrate , n = Molarity x volume in L

                                                  = 0.40 M x ( 15.0 mL x (1L/1000mL) )

                                                 = 0.006 moles

The balanced equation is   : NaCl + AgNO₃ Ag + NaNO₃

According to the balanced equation ,

1 mole of 1 mole of NaCl reacts with 1 moles of AgNO₃

0.006 mole of 1 mole of NaCl reacts with 0.006 moles of AgNO₃

So the number of moles of HCl required is 0.006 moles

Given Molarity of HCl , M = 1.30 M

We know that Molarity , M = Number of moles / volume in L

So Volume of HCl in L , V = number of moles / Molarity

                                     = 0.006 mol / 1.30 M

                                    = 4.6x10^-3 L

                                    = 4.6 mL               Since 1 L = 10^-3 mL

Therefore the volume of HCl required is 4.6 mL