Question

In the laboratory a student combines 43.5 mL of a 0.111 M iron(II) acetate with 10.7 mL of a 0.359 M chromium (III) acetate solution

what is the final concentration of acetate anion?

if you could explain how to get the answer in a visual, that would be helpful.

Answer 1

0.3908M

Explanation

Concentration of iron(ll)acetate = 0.111M

0.111M = 0.111mol in one liter ( 1mole = 6.022×10^23 atom/molecule/ion)

Volume of iron(II) acetate = 43.5ml

Therefore,

No of mole of iron acetate present = (0.111mole/1000ml)× 43.5ml = 0.0048285

1 molecuele of Iron(II) acetate( Fe(CH3COO)2) contain 2mole

of CH3COO-

Therefore,

No of mole of CH3COO- by Iron(II) acetate = 2× 0.0048285mol = 0.009657

similarly,

Concentration of Chromium(III) acetate = 0.359M

0.359M = 0.359 mole in one liter

Volume of Chromium(III)acetate solution = 10.7ml

Therefore,

No of mole of Chromium(III)acetate =

( 0.359mol/1000ml)×10.7ml = 0.0038413

1molecule of Chromium(III)acetate ( Cr(CH3COO)3) containe 3 CH3COO- ions

Therefore,

No of mole of CH3COO- ion present in Chromium(III)actate solution = 3× 0.0038413= 0.011524

Now,

Total no of mole of CH3COO- = 0.009657 + 0.011524=0.021181

Total volume = 43.5ml + 10.7ml = 54.2ml

Therefore,

Concentration of acerate ion = (0.021181mol/54.2ml)×1000ml = 0.3908M