Question

Solid sodium iodide is slowly added to 175 mL of a 0.468 M silver acetate solution until the concentration of iodide ion is 0.0687 M. The mass of silver ion remaining in solution is _______  grams

Solid chromium(III) acetate is slowly added to 75.0 mL of a 0.388 M potassium hydroxide solution until the concentration of chromium(IIID ion is 0.0556 M. The percent of hydroxide ion remaining in solution is _______ % 

Answer 1

Answer – We are given, [AgCH₃COO] = 0.468 M , volume = 175 mL

[NaI] = [I^-] = 0.0687 M

Reaction - AgCH₃COO + NaI -----> AgI + CH₃COONa

There is mole ratio 1:1

We need to calculate the moles of each AgCH₃COO and NaI

Moles of AgCH₃COO = 0.468 M x 0.175 L

                                      = 0.0819 moles

Moles of I^- = moles of NaI = 0.0687 M x 0.175 L

                                               = 0.0120 moles

Now we know there is mole ration 1:1 and moles of I^­­- is lowest, so NaI is limiting reactant

So mole of AgCH₃COO used = 0.0120 moles

Moles AgCH₃COO remaining = 0.0819 moles – 0.0120 moles

                                                   = 0.0698 moles

Moles of AgCH₃COO = Ag⁺ moles of = 0.0698 moles

Mass of Ag⁺ remaining = 0.0698 moles x 107.86 g/mol

                                          = 7.54 g

We are given, [KOH] = 0.388 M , volume = 0.075 L , [Cr³⁺] = 0.0556 M

Reaction - Cr(CH₃COO)₃ + 3 KOH -----> Cr(OH)₃ + 3 CH₃COOK

First, we need to calculate the moles KOH

Moles of KOH = 0.388 M x 0.075 L

                          = 0.0291 moles

moles of KOH = moles of OH^- = 0.0291 moles

Moles of Cr³⁺ ion = 0.0556 M x 0.075 L

                                 = 0.00417 moles

We are given to calculate the excess mass of hydroxide ion, means limiting reactant is Cr(CH₃COO)₃

moles of KOH used –

1 mole of Cr(CH₃COO)₃ = 3 moles of KOH

0.00417 moles of Cr(CH₃COO)₃ = ? moles of KOH = ?

= 0.0125 moles of KOH used

moles of KOH remaining = 0.0291 – 0.0125

                                             = 0.0166 moles

moles of KOH = moles of OH^­- = 0.0166 moles

Percent of OH^- ion remaining = 0.0166 / 0.0291 x 100 %

                                                      = 57.0 %