Question

1. Moles of AgOH produced

1.0mL of 0.03 M AgNO3 + 1.0mL of 0.03 M NaOH

2. Volume of 1.0 M NH3 required to react with all the AgOH produced

1.0 M NH3 used

3. Assuming one drop has a volume of 0.05mL, what is the minimum number of drops of 1.0 M NH3 required.

Answer 1

Reactions of AgNO₃ with NaOH is a follows :-

Step 1: Aqueous silver nitrate is mixed with aqueous sodium hydroxide.

   AgNO3   + NaOH → AgOH   + NHO3        -----   equation (1)

AgOH is thermodynamically unstable in solution and decomposes into Ag₂O ( a dark brown precipitate is formed).

2AgOH    →   Ag2O + H2O                         --------- equation (2)

Step 2: Aqueous ammonia is added drop-wise until the precipitated silver oxide completely dissolves.

Ag2O   +   4NH3+ H2O   →    2[Ag(NH3)₂]⁺ +   2OH−           ----- equation (3)

    

                               2[Ag(NH3)₂]⁺    ==         (silver diamine complex or tollen's reagent)

Now,

number of moles = Volume (in Litres) * Molarity(M)

number of moles of AgNO₃ = 0.03 M * (10^-3) L = 0.03 * 10^-3 moles or 0.03 milli moles

number of moles of NaOH = 0.03 M * (10^-3) L = 0.03 m moles     {m moles = milli moles}

(i)

1 moles of AgNO₃     forms 1 moles of AgOH   (from equation 1)

so , 0.03 m moles of AgNO₃   ==> 0.03 m moles of AgOH

Moles of AgOH produced = 0.03 m moles

(ii)

similary from equation 2

2 moles of AgOH     forms 1 moles of Ag₂O

so , 0.03 m moles of AgNO₃   ==> 0.015 m moles of Ag₂O   

(iii)

from equation 3

1 moles Ag₂O react with 4 moles of NH₃

so, 0.015 m moles Ag₂O react with 4*0.015 m moles of NH₃ = 0.06 m moles

volume[in Litres] of NH₃ required = (number of moles) / (Molarity )          molarity = 1.0M (given)

volume[in Litres ]of NH₃ required = (0.06 * 10^-3)/1 = 0.06*10^-3 L

Volume of NH₃ required = 0.06 mL

number of drops required = (Vol of NH₃ ) / (vol of each drop) = 0.06 / 0.05

number of drops required = 1.2 drops

So , the minimum number of drops required to completely react with Ag₂O will be 2 .