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12. Titrations

posted Apr 4, 2016, 3:32 AM by Upali Salpadoru   [ updated Jun 2, 2017, 12:53 PM ]

     

 

                                            

Strong acids and caustic alkalis can cause chemical burns.
 Although one can kill the other, in case of an accident you should not use the other, as the resulting reaction would be too vigorous and exothermic   ( heat producing).
The best thing would be to softly flush the affected area with tap water for about 10 minutes.
 

Testing concentrations.                        

Titration is the method used to determine the strength of a solution using a solution of known strength. The apparatus used for this purpose is depicted above.

Very often this is employed for acid base reactions. If you wish to determine the strength of a sodium hydroxide solution, you can take a measured olume into a titration flask. The burette can be filled with the acid of known strength. We have to add an indicator to find the exact point of neutralization.

                   Acid and Base reactions

               Consider this reaction:     NaOH + HCl  =   NaCl  + H2O

              Mole ratio:- …………      1mol    :    1mol

              This shows that 1mol of acid can completely neutralise 1 mol of the base.

                  This shows that 1.L of  1M-NaOH can be completely neutralised by 1L of 1M – HCl.

 

When the reacting mole ratio is 1:1,

equal volumes of identical strength can neutralize each other.

                      Then the formula is:-         

                 

MaVa    Product of  molarity and     volume of acid.

MbVb   Product of molarity and volume of base

Example 1.

25 ml of  a   0.1 M- sodium hydroxide solution was completely neutralised by  21.2 ml of  HCl.  Find the strength of the acid.

Equation:-   NaOH  + HCl  ------>  NaCl  +  H2O.

Molar ratio:-     1     :      1

                           x  V a    =  M x V

                  M a   x  21.2         =      0.1 x  25

                     M a       …        =  0.1 x25/ 21.2  =  0.12 M              

We may consider this reaction in the ionic form.

What actually happens is this:-     

                   NaOH  <-------> ionises as  ( Na+ )  +  ( OH-

                    HCl    <------->  ..................( H+ )     +   (Cl-)


While sodium ions and chloride ions remain in the solution as they are hydrogen ions combine with hydroxyl ions forming water molecules.

So simply the reaction is :-

                       H+ +  OH-    ------->    H2O

Ratio :              1    :    1


         When the concentrations of the two solutions are not the same we cannot use this                   formula:-

            In that case the formula has to be :- 

          

Ra: Rb  is the neutralizing ratio of acid and base.

                   


                      Example 2

                               2NaOH  + H2SO4  =   Na2SO4  + 2H2O

                 Mole ratio:   2     :     1.       

                   Sodium hydroxide volume -   1000 ml      strength    2M

Text Box


                   Sulphuric acid        volume       Va              strength   3M

                   Find the volume of sulphuric acid required.

 

                  Answer

                   No. of moles in 1.L of NaOH……2

                   H2SO4 moles required for 2 mol. of NaOH  ..1

                    Volume of H2SO4   that has 3 mol.        1.L

               Volume of H2SO4   ……………1 mol            = 1/3  L  =   333.33 mL.


                  Answer using Formula                              

                           M a  x  V a    M b x  Vb 

                                      EMa                     EMb

                          3 x V           =   2 x 1000

                                1                       2

                                    V      =   2000/6  =    333.33 mL.

        Example 3

             Sodium carbonate  =  Mass ….- 5.3g.    Volume 250 mL.

             HCl required to neutralise =    Volume  30 mL.

             Find the strength of the acid solution.

          

        Answer

                 Equation:-  Na2CO3 +2 HCl  ----à 2 NaCl  +  H2CO3

        First of all we have to find the molarity of the base.

          Mass dissolved in 250 mL =  5.3g

          Mass……………1000 mL =  5.3x4  = 21.2 g

          Converting this to moles     =  21.2/106 = 0.2moles

          Formula                           M a  x  V a    M b x  Vb 

                                                                 EMa                   EMb

          Substitution:                   M x 30  =  0.2 x 25

        Mole ratio                2        :         1

                                        M =   0..2x25x2/ 30  =0.333

         Eample 4

         2M Hydrochloric acid ,100 mL, was added to 10g of  mortar containing lime and sand  Assume that CaO was the only reactant present . The remaining solution was filtered and titrated with a 2M –NaOH solution. The volume of alkali required was  75.0 mL.  Find the percentage of CaO in mortar.

                   First equation.             CaO  + 2 HCl  ----à   CaCl2  +  H2O

                       Second equation              NaOH +  HCl  ----à    NaCl  +  H2O

                 We have to work on the second equation first.        

                       As the mole ratio is 1:1 we can use   M a  x  V a    M b x  Vb 

                                                                                                 2   x V         =   2 x 75

                                                                                                 V  =   75 mL of acid,

                     Volume of acid added to mortar  =100 mL

                            Volume of acid reacted                 =  100-75 = 25 mL

                            No. of moles of acid in 25 mL      2x25/1000  =  50/1000   = 1/20  =0.05

                                    CaO  + 2 HCl  ----à   CaCl2  +  H2O

              Mole ratio         1     :      2.  

              2 mol of acid can react with       1 mol of base,

               0.05…………………………..1/2 x0.05  =   0. 0 25 mol.

                    Mass of 1 mole of CaO  =   56 g

                                  0.025mol          =  56 x0.025 =  1.4g

              1.4g present in 10 g of mortar.

               Mass of CaO in   10g  of mortar ….=   1.4 g

               Mass…………………….100g……=    14g.

              As a  Percentage of CaO           = 14 %.   Of CaO.

                 

               1.0 

Write the equation for each reacti0n and determine what is not given.

Acid  strength

Acid Volume

Base  strength

Base volume

1….    HCl + NaOH  ------> NaCl  +  H2O

      1 M- HCl

25 mL

1M =NaOH

i.………….

2.

     2M - HCl

25 mL

1 M -NaOH

ii…………...

3.   H2SO4  +   2 NaOH ------->   Na2SO4  +  2 H2O

    1M= H2SO4

25 mL

1 M - NaOH

iii………………...

4.  2M=H2SO4


25 mL

5 M- NaOH

iv…………….

5.      HNO3   +   KOH  ------>     KNO3  + H2O

  2M-HNO3

10 mL

V,,,,,, KOH

50 mL

Answers

I.  25 mL.   Ii. 50 mL.  Iii. 50 mL.  Iv.  20 mL  v,  0.4 M.



                   2.0  Find the answers for these.                       

i. Give the reacting molar ratios for the reactants and products for this equation:

       2 HCl  + K2CO3 ------> 2 KCl  + H20  + CO2

                  ii. What is the volume of 1 M- HCl required to react with 0.5 mol of K2CO3 

 iii. If  0.05 L of  potassium carbonate produced a residue of  5.8g on evaporation what would have been the molarityof the solution?

iv.  If    18.5  mL  of HCl could neutralise   25 mL of 0.5 M  potassium carbonate solution, what is the strength of the acid?

v. How many moles of CO2 will be produced by the above reaction?

ANSWERS

i.        2: 1 ------>  2 : 1:1

ii.       To react with 1 mol. of   K2CO3 volume of acid required = 2 mol.

                               0.5 mol......................................................     = 1 mol.

iii. 

             Mass of solute in 0.05 L of solution  =  5.8g  
             Mass ................... 1 L  .....................=   5.8/ 0.05 =  116g.
              Converting 116 g to moles .............=116/138.2  = 0.84  mol.
               Molarity of solution ................      = 0.84 M.
              
                             

                   3.0

                   Fill in the blank spaces in this table so as to neutralise each other.

                        Equation:-          KOH  +   HCl -----à   KCl  +  H2O

             

Acid Volume

Acid strength

Base volume

Base strength

20mL

1-M.

 10 mL

i.                  …………….

 12 mL

2-M

ii ……

 3-M

15 .5 mL

 iii…………

20 mL

 0.5 M.

 iv…………

 0.1 M

 0.25 L 

  0.1 M

 25mL

v…………..

 22 mL

5.6g/L

                                                                                                                         

                4.0    This question is about acetic acid in vinegar.

                 i.       What is the molar mass of this acid.

                ii.      A sample of home vinegar has 4% acetic acid.      Assuming 100g of water to have 4g of vinegar find the molarity of the sample..

                iii.     How many grams of CO2 can you obtain by the reaction of 25mL of this vinegar on baking powder.

            Word equation:-    

            Sodium bi carbonate  + acetic acid -------à  sodium acetate  + water + carbon dioxide.

            Equation:- ……………….NaHCO3 + CH3COOHCH3COONa + H2O + CO2

        Answers 

          i.   CH3COOH    = 12+3+12+32+1 = 60g/mol

          ii.:  As the density of water is 1g /cm3, the volume of 100g of water = 100ml.

                                                 100mL  will have                   4g of acid.

                                 1000 mL……………………..40g.

                                                  Converting this to moles   ,,,  =  40/60  = 2/3  =  0.67 M     

 

           iii.

                 Equation:                          NaHCO3 + CH3COOHCH3COONa + H2O + CO2

                 Molar ratio:                            1          :        1                                                                1       

                                               1000 ml acid solution      has    0.67 mol

                                            25 mL………………………..= 0,67 x 25/1000  =0.017 mol

                                                CO2 produced by this will be same number of moles  = 0.017 mol

                                                1 mol of CO2 ……………………….44g

                                                 0.017 mol………………………….= 44x0.017 =  0.75g.

    


Write the equation for each reaction and determine what is not given.

Acid  strength

Acid Volume

Base  strength

Base volume

1….    HCl + NaOH  ------> NaCl  +  H2O

      1 M- HCl

25 mL

2M =NaOH

i.………….

2.

     2M - HCl

25 mL

1 M -KOH

ii…………...

3.   H2SO4  +   2 NH4OH ------->   (NH4)2SO4  +  2 H2O

    1M= H2SO4

25 mL

1 M - NH4OH

iii………………...

4.  3M=H2SO4


25 mL

5 M- NaOH

iv…………….

5.      HNO3   +   KOH  ------>     KNO3  + H2O

  0.5 -M-HNO3

10 mL

V,,,,,, KOH

50 mL



1.0 This question is about sodium carbonate and hydrochloric acid.

Equation:-      Na2CO3   +   2 HCl   ------->  2 NaCl   +  H20

A decimolar solution (0.1-M)  of sodium carbonate was titrated to find the strength of hydrochloric acid.   0.25 L of Na2CO3 required 0.20.1 L of acid.

 Find the following:

 i.  Molar massof sodium carbonate.

ii.  No. of moles in 0.25  L. of sodium carbonate solution.

iii. No of moles of HCl that the abobe 0.25 L can neutralize.

iv. The strength of the acid solution.                                                                                                               

 2.0  This question is about magnesium hydroxide. Milk of magnesia ,Mg(OH)2 is a mild    alkali that was used to neutralize excessive hydrochloric acid in the stomach.                               i.     What is the molar mass of Mg(OH)2  /               

        ii.   This substance was used to reduce acidity in the stomach. Acid molarity in         digestive juice can go up to o.1M. of HCl. Find the mass of Mg(OH)2  that can neutralize acids in 20 mL of juice                                                        

                                                                                     (5x2=10 points)

3.0

25 mL of Phosphoric acid was used to react with 7.4 g of calcium hydroxide .
 Equation:    3 Ca(OH)2  +2  H3PO4------à   Ca3(PO4)2 + 6 H2O
i.            Molar mass of the acid.
ii.           No. of Ca(OH)2   moles in 7.4 g.
iii.         Molarity of acid
iv.         Mass of calcium phosphate formed.

                                                                            (5x4= 20)

 


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