Question

One deciliter of a .9% nickel (II) nitrate solution is diluted to a volume of 500 mL. Calculate the final [NO3-] of the resultant solution. Answer in M please.

Answer 1

1 deciliter = 100 ml

first calculate final concentrarion of nickel (II) nitrate solution

Use the formula C₁V₁ = C₂V₂

Where C₁ = initial concentration = 0.9%

V₁ = initial volume = 100 ml

C₂ = final concentration = ?

V₂ = final concentration = 500 ml

Substitute the value in formula

C₁V₁ = C₂V₂

C₂ = C₁V₁/V₂

= 0.9 100/500

C₂ = 0. 18 %

structure of nickel (II) nitrate = Ni(NO₃)₂

nickel (II) nitrate dissociate as  Ni(NO₃)₂  Ni²⁺  + 2 NO₃^-

thus concentration of NO₃^- is double than Ni(NO₃)₂

[NO₃^-] = 2 0.18 = 0.36 %

Conversion % to molar

you have not given density of solution so consider density equal to water = 1 then 500 ml solution = 500 gm solution.

calculate mass of NO₃^-

mass of NO₃^-  = mass of soution %mass of NO₃^- / 100 = 500 0.36 /100 = 1.8 gm

molar mass of  NO₃^- = 62 gm/mol then 1.8 gm  NO₃^-  = 1.8/62 = 0.029 mole

molarity = no.of mole/volume of solution in liter

molarity of  NO₃^-  = 0.029/0.5 = 0.058 M

[ NO₃^-] = 0.058 M