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Equilibrium

Question
CBSEENCH11006561
Wired Faculty App

What is the pH of 0·001M aniline solution? The ionisation constant of aniline is 4·27 × 10–10. 
Calculate the degree of ionisation of aniline in the solution. Also,calculate the ionisation constant of the conjugate acid of aniline.

Solution
straight C subscript 6 straight H subscript 5 NH subscript 2 space plus space straight H subscript 2 straight O space space space rightwards harpoon over leftwards harpoon space space space space space straight C subscript 6 straight H subscript 5 straight H subscript 3 superscript plus space plus space OH to the power of minus space space space straight K subscript straight a space equals space fraction numerator open square brackets straight C subscript 6 straight H subscript 5 NH subscript 3 superscript plus close square brackets space open square brackets OH to the power of minus close square brackets over denominator open square brackets straight C subscript 6 straight H subscript 5 NH subscript 2 close square brackets end fraction Here space open square brackets straight C subscript 6 straight H subscript 5 NH subscript 3 superscript plus close square brackets space equals space open square brackets OH to the power of minus close square brackets therefore space space space space space space straight K subscript straight a space equals space fraction numerator open square brackets OH to the power of minus close square brackets squared over denominator open square brackets straight C subscript 6 straight H subscript 5 NH subscript 2 close square brackets end fraction or space space space space space space open square brackets OH to the power of minus close square brackets space equals space square root of straight K subscript straight a space cross times space open square brackets straight C subscript 6 straight H subscript 5 NH subscript 2 close square brackets end root or space space space space space space space open square brackets OH to the power of minus close square brackets space equals space square root of left parenthesis 4.27 space cross times space 10 to the power of negative 10 end exponent right parenthesis thin space left parenthesis 10 to the power of negative 3 end exponent right parenthesis end root space space space space space space space space space space space space space space space space space space space space equals space 6.534 space cross times space 10 to the power of negative 7 end exponent straight M

But comma space pOH space equals space minus log space open square brackets OH to the power of minus close square brackets space space space space space space space space space space space space space space equals space minus space log space left parenthesis 6.534 space cross times space 10 to the power of negative 7 end exponent right parenthesis space space space space space space space space space space space space space space equals space 7 minus 0.8152 space equals space 6.18 therefore space space space space space space space pH space equals space 14 space minus space pOH space equals space 14 minus 6.18 space equals space 7.82

left parenthesis ii right parenthesis space Also comma space space space space space space space space space straight C subscript 6 straight H subscript 5 NH subscript 2 space plus space straight H subscript 2 straight O space space space space space rightwards harpoon over leftwards harpoon space space space space straight C subscript 6 straight H subscript 5 NH subscript 3 superscript plus space plus space OH to the power of minus Initial space space space space space space straight C At space eqm. space space space straight C space minus Cα space space space space space space space space space space space space space space space space space space space space space space Cα space space space space space space space space space space space space space Cα space space space space space space space space

therefore space space space straight K subscript straight b space equals fraction numerator open square brackets straight C subscript 6 straight H subscript 5 NH subscript 3 superscript plus close square brackets space open square brackets OH to the power of minus close square brackets over denominator open square brackets straight C subscript 6 straight H subscript 5 NH subscript 2 close square brackets end fraction space equals space fraction numerator Cα space cross times space Cα over denominator straight C left parenthesis 1 minus straight alpha right parenthesis end fraction space space space space space space space space space space equals space fraction numerator Cα squared over denominator 1 minus straight alpha end fraction space asymptotically equal to space space Cα squared therefore space space space space space straight alpha space equals space square root of straight K subscript straight b over straight C end root space equals space square root of fraction numerator 4.27 space cross times space 10 to the power of negative 10 end exponent over denominator 10 to the power of negative 3 end exponent end fraction end root space space space space space space space space equals space 6.53 space cross times space 10 to the power of negative 4 end exponent

Also comma space space space pK subscript straight a space plus space pK subscript straight b space equals space 14 space left parenthesis for space straight a space pair space of space conjugate space acid space and space base right parenthesis Now comma space space space space pK subscript straight b space equals space minus log space straight K subscript straight b space equals space minus log space left parenthesis 4.27 space cross times space 10 to the power of negative 10 end exponent right parenthesis space space space space space space space space space space space space space space space space space equals space minus 0.62 space plus space 10 space equals space 9.38 therefore space space space space space pK subscript straight a space equals space 14 minus pK subscript straight b space equals space 14 minus 9.38 space equals space 4.62 straight i. straight e. space space space minus log space straight K subscript straight a space equals 4.62 or space space space space space space space space log space straight K subscript straight a space equals space minus 4.62
equals space minus 4.0 space minus space 0.62 space plus space 1 space minus 1 space equals space 5 with bar on top.38
therefore space space space straight K subscript straight a space equals space antilog left parenthesis 5 with bar on top space.38 right parenthesis space equals space 2.399 space cross times space 10 to the power of negative 5 end exponent space space space space space space space space space space space equals 2.4 space cross times space 10 to the power of negative 5 end exponent

 

Some More Questions From Equilibrium Chapter

Give two examples of physical equilibrium.

What kind of molecules in a liquid can evaporate?

Name the factors on which vapour pressure of any liquid depends.

Which measurable property becomes constant in water vapour equilibrium?

Give an example from daily life in which there is gas solution equilibrium?

A crystal of common salt of a given mass is kept in its aqueous solution. After 24 hours, its mass remains the same. Is the crystal in equilibrium with the solution?

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