Question

Calculate the value of the reaction quotient, Q, of silver iodate (AgIO₃) when 10.0 mL of 0.018 M AgNO₃ is mixed with 10.0 mL of 0.018 M NaIO₃. (Please give your answer with 2 significant figures.)

Answer 1

10.0 mL of 0.018 M AgNO₃ 

10.0 mL of 0.018 M NaIO₃.

Reactions are:-

AgNO_3(aq)+NaIO_3(aq)$\to$ AgIO_3(s)+NaNO_3(aq)

AgIO_3(s)$\leftrightarrow$ Ag⁺_(aq)+IO₃^-_(aq)

Q(Reaction quotient) =[Ag⁺][IO^-₃]

Moles of AgNO₃=Molarity ×Volume

$\frac{10.0 mL ×0.018 mol/L}{1000mL/L}=$ 1.8×10^-4moles

Using moles ratio,1:1

AgNO₃:NaIO₃

Moles of NaIO₃=1.8×10^-4moles

Total volume after mixing =10mL+10mL=20mL

Concentration of [Ag⁺]=

$\frac{1000mL/L}{20mL}$ ×1.8×10^-4mol

=9×10^-3M

Using mole ratio:-

[Ag⁺]:[IO^-₃]

1:1

Therefore the concentration of [IO^-₃]

=9×10^-3M

Reaction quotient(Q)=[Ag⁺][IO^-₃]

=[9×10^-3][9×10^-3]

=81×10^-6

Q=8.1×10^-5