Calculate the concentrations of iron(iii) ions and scn- ions when 5.00 ml of 0.002m iron(iii) nitrate is mixed with 4.00 ml of 0.002 m nascn and dilute nitric acid is added to bring the total volume of the solution to 20.0 ml. what is the concentration of iron(iii) ions?

Volume of Fe(NO₃)₃ = 5.00 ml

Total voluem of the solution = 20.0 ml

Concentration of Fe(NO₃)₃ = 0.002 M

Assuming complete dissociation of  Fe(NO₃)₃ concentration of Fe³⁺  = 0.200 M

Using the dilution equation, M₁V₁ = M₂V₂

where M₁ and M₂ are the initial and final concentration of Fe³⁺  respectively and V₁ and V₂ are the initial and final volume of Fe³⁺  respectively

Plugging the data we get,

M₂ = = 5 x 10⁻⁴

Therefore, the concentration of Fe³⁺ once diluted in the mixture is 5 x 10⁻⁴M  

its first and third is your answer i did the qustion the fourth isnt right

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.