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CBSE Class 11-science Answered

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Asked by Prashant DIGHE | 13 Jun, 2020, 10:21: PM
answered-by-expert Expert Answer
71. Neutralisation equation-
 HCl plus NaOH rightwards arrow NaCl space plus straight H subscript 2 straight O
straight n equals 100 cross times 1 equals 100 space straight m space mole equals 0.1 space mole

Energy space evolved space due space to space neutralization space of space HCl space and space NaOH space equals 0.1 cross times 57 equals 5.7 space KJ equals 5700 space Joule
Energy space used space to space increase space temperature space of space solution equals 200 cross times 4.2 cross times 5.7 equals 4788 space Joule
Energy space used space to space increase space temperature space of space calorimeter equals 5700 minus 4788 equals 912 space Joule
ms. increment straight T equals 912
ms cross times 5.7 equals 912
ms equals 160 space Joule divided by ring operator
Energy space evolved space by space neutralization space of space CH subscript 3 COOH space and space NaOH
equals 200 cross times 4.2 cross times 5.6 plus 160 cross times 5.6 equals 5600 space Joule

so comma space energy space used space in space dissociation space of space 0.1 space mole space CH subscript 3 COOH space equals 5700 minus 5600 equals 100 space Joule
Enthalpy space of space dissociation space equals 1 space KJ divided by mole


72.
left square bracket CH subscript 3 COOH right square bracket equals fraction numerator 1 cross times 100 over denominator 200 end fraction equals 1 half
left square bracket CH subscript 3 CONa right square bracket equals fraction numerator 1 cross times 100 over denominator 200 end fraction equals 1 half
pH equals pK subscript straight a plus log fraction numerator left square bracket Salt right square bracket over denominator left square bracket Acid right square bracket end fraction
pH equals 5 minus log 2 plus log fraction numerator 1 half over denominator 1 half end fraction
pH equals 4.7
Answered by Ravi | 14 Jun, 2020, 06:45: PM
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