Question

500 ml of 2M HCl,100ml of 2M H2SO4 and one gram equivalent of monoacidic alkali are mixed together.30ml of this solution required 20ml of 143gm Na2CO3.xH2O in one litre solution. Calculate water of crystallisation of Na2CO3.xH2O.

Answer 1

 The water of crystallization will be 10H2O....

500 ml of 2M HCl +100ml of 2M H2SO4 

Total milliequivalents of Hydronium ion will be==>>500x2+100x4{since H2SO4 will give 2 H+}

==>>1400milliequivalents..=1.4equivalents

The equivalents of OH- added in form of Monobasic alkali=1

Net after neutralization will be 0.4 equivalents of Hydronium ion.

30ml of this solution required 20ml of 143gm Na2CO3.xH2O in one-litre solution.

Now just equate both milliequivalents....

We know 2H+ + Na2CO3===>  2NaOH + CO2

THAT MEANS 2 HYDRONIUM IONS ARE NEEDED FOR EACH Na2CO3 MOLECULE

n factor for Na2CO3 is 2

Molarity of Na2CO3=$\frac{143}{106+18x}$

Normality=Mxn factor =2M

0.4x30mL=   2$\frac{143}{106+18x}$  x20mL

x=10

Answer 2

• Answer: 10  Explanation: The water of crystallization will be 10H2O....  500 ml of 2M HCl +100ml of 2M H2SO4   Total milliequivalents of Hydronium ion will be==>>500x2+100x4{since H2SO4 will give 2 H+}  ==>>1400milliequivalents..=1.4equivalents  The equivalents of OH- added in form of Monobasic alkali=1  Net after neutralization will be 0.4 equivalents of Hydronium ion.  30ml of this solution required 20ml of 143gm Na2CO3.xH2O in one-litre solution.  Now just equate both milliequivalents....  We know 2H+ + Na2CO3===>  2NaOH + CO2  THAT MEANS 2 HYDRONIUM IONS ARE NEEDED FOR EACH Na2CO3 MOLECULE  n factor for Na2CO3 is 2  Molarity of Na2CO3= 143/(106+18x)  Normality=Mxn factor =2M  0.4x30ml = 2(Molarity)×20ml  x=10  ANSWER