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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

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Sulphuric acid        volume       Va              strength   3M

Find the volume of sulphuric acid required.

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.

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.

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

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?

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

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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