Respuesta :
Answer:
The answer is 465.6 mg of MgI₂ to be added.
Explanation:
We find the mole of ion I⁻ in the final solution
C = n/V -> n = C x V = 0.2577 (L) x 0.1 (mol/L) = 0.02577 mol
But in the initial solution, there was 0.087 M KI, which can be converted into mole same as above calculation, equal to 0.02242 mol.
So we need to add an addition amount of 0.02577 - 0.02242 = 0.00335 mol of I⁻. But each molecule of MgI₂ yields two ions of I⁻, so we need to divide 0.00335 by 2 to find the mole of MgI₂, which then is 0.001675 mol.
Hence, the weight of MgI₂ must be added is
Weight of MgI₂ = 0.001675 mol x 278 g/mol = 0.4656 g = 465.6 mg
We have the milligrams of MgI2 must be added to 257.4 mL of 0.078 M KI to produce a solution with [I−] = 0.1000 M is
[tex]m_{MgI2 }=465.65mg[/tex]
From the Question we are told that
volume of solution =257.7 [tex]ml =0.2577L[/tex]
Initial conc of [tex][I^-][/tex] [tex]C_1 = 0.087M[/tex]
Final conc of [tex][I^-][/tex] [tex]C_2= 0.1M[/tex]
Generally
Net Concentration of [I^-]
[tex]C'=C_2-C_1\\\\C'= 0.1-0.087\\\\C'=0.013mole/L[/tex]
no of moles of I^- required
m = molarity* volume of solution in L
[tex]m = 0.013*0.2577\\\\m= 0.00335 moles of I^-[/tex]
With ratio of MgI_2 to I^- as 1:2
Therefore
The no of moles of MgI2 needed = 0.00335/2
The no of moles of MgI2 needed =0.001675
Hence
mass of MgI2
[tex]m_{MgI2 }=[/tex] no of moles * gram molar mass
[tex]m_{MgI2 }=0.001675*278[/tex]
[tex]m_{MgI2 }=0.46565g[/tex]
[tex]m_{MgI2 }=465.65mg[/tex]
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