TUGAS INDIVIDU KIMIA ANALITIK "TITRASI"

1.       50,00 mL of  HCl was titrated with 0,01963M Ba(OH)_2. The end point was reached (using bromocresol green as indicator after 29,71ml Ba(OH)₂ was added. What is the concentration of the HCl ?  

Answer:

Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O

1mol  Ba(OH)₂  reacts  with 2mmol of  HCl

C HCl .V HCl       =              C _Ba(OH)2  . V _Ba(OH)2  

      n HCl                                              n _Ba(OH)2  

C HCl . 50,00  =              0,01963M  . 29,71ml  

         2                                                        1  

C HCl =   (29,71 x 0,01963 x 2) / 50,00

        =  1,1664146  / 50,00

        =  0,023328292 M

        = 0,02M

2.      A 0,8040 g sample of an iron are is dissolved in acid. The iron is then reduced to Fe³⁺ and titrated with of 0,02242 M  KMnO₄,  47,22 mL of titrant was added to reach the end point. Calculate the % Fe in the sample!

(BM =55.847 g/mol).

MnO₄ ^- + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O

Answer:

 stoichiometric ratio =  5 mmol Fe ²⁺

                                       1 mmol KMnO₄

amount KMnO₄ = 47,22 ml × 0,02242 mol/L = 1,0587 mmol

amount Fe²⁺ = 47,22 ml × 0,02242 mol/L × 5 = 5,2934 mmol

mass Fe²⁺ = 5,2934 mmol × 0,055847 g/mmol = 0,2956 g

percent Fe²⁺ =  (47,22  x 0,02242  x 5 x 0,055847) g    x  100% 

                                                0,8040 g sample

         = 36.77%

3.      The CO in a 20,3L sample of gas was converted to CO₂ by passing the gas over iodine pentoxide heated to 150℃,

I₂O_{4   (s)} + 5CO  _(g) → 5CO_{2  (g)} + I_{2  (g)}         

The iodine distilled at this temperature was collected in an absorber containing 8,25 ml of  I₂ was distilled and collected to 8.25 mL of   0.01101 M

I_{2  (aq)} + 2S₂O₃^2-  _(aq) → 2I^-  _(aq) + S₄O₆^2-  _(aq)    

The excess Na₂S₂O₃ was back titrated with  2,16 ml of 0,00947 M  I₂ solution. Calculate the mg of CO per liter of sample. (BM_CO =28.01 g/mol)

Answer:

I₂O_{4   (s)} + 5CO  _(g) → 5CO_{2  (g)} + I_{2  (g)}          [CO : I₂ = 5 : 1]

I_{2  (aq)} + 2S₂O₃^2-  _(aq) → 2I^-  _(aq) + S₄O₆^2-  _(aq)   [I₂ : S₂O3^2- = 1 : 2]

[CO: S2O3^2- = 5 : 2]

Mass CO =  (8,25× 0,01101- 2,16× 0,00947 × 2)× ⁵/₂  x 28,01

                                                20,3 L

                            =  0,12480525 mmol x 28,01     

                                                20,3                                 

                            =   3,495795053     

                                    20,3

    =  0,1722066528  mg/L

    =  0,17 mg/L