Question

Exercise 18.108 A 130.0 −mL sample of a solution that is 3.0×10−3 M in AgNO 3 is mixed with a 225.0 −mL sample of a solution that is 0.13 M in NaCN . Part A After the solution reaches equilibrium, what concentration of Ag + (aq) remains? Express your answer using two significant figures

Answer 1

Number of moles of AgNO₃ is , n = Molarity * volume in L

                                                   = 3*10 ^-3 M * 0.130 L

                                                     = 0.00039 moles

Number of moles of NaCN is , n' = molarity * volume in L

                                                    = 0.13 M * 0.223 L

                                                    = 0.0289 moles

AgNO₃ + NaCN ----> AgCN + NaNO₃

According to the balanced Equation ,

1 mole of AgNO₃ reacts with 1 mole of NaCN

0.00039 moles of AgNO₃ reacts with 0.00039 moles of NaCN

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1 mole of AgNO₃ produces 1 mole of AgCN

0.00039 moles of AgNO₃ produces 0.00039 moles of AgCN

Total volume of the solution is = 130 mL + 225 mL

                                              = 355 mL

                                              = 0.355 L

So total equilibrium concentration of AgCN is = No.of moles / volume in L

                                                                      = 0.00039 moles / 0.355 L

                                                                      = 0.001098 M

AgCN ---> Ag ⁺ + CN^-

1 mole of AgCN contains 1 mole of Ag+

So concentration of Ag⁺ = concentration og AgCN + concentration of AgNO₃ left

                                      = 0.001098 M + 0 M

                                      = 0.001098 M