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Equilibrium

Question
CBSEENCH11006614
Wired Faculty App

Determine the solubility of ferric hydroxide at 298K. Given Ksp for Fe(OH)3 = 1·0 × 10–38. Also, determine the molarities of each ion. 

Solution
Let the solubility of Fe(OH)3 in water be S mol–1  
  Fe left parenthesis OH right parenthesis subscript 3 space space space space space rightwards harpoon over leftwards harpoon space space space space space Fe to the power of 3 plus end exponent left parenthesis aq right parenthesis space plus space 3 OH to the power of minus left parenthesis aq right parenthesis straight i. straight e. space 1 space mole space of space Fe left parenthesis OH right parenthesis subscript 3 space in space the space solution space gives space 1 space mole space of space Fe to the power of 3 plus end exponent space ions space and space 3 space moles space of space OH to the power of minus space ions Hence space in space solution open square brackets Fe to the power of 3 plus end exponent close square brackets space equals space straight S space mol space straight L to the power of negative 1 end exponent semicolon space space open square brackets OH to the power of minus close square brackets space equals space 3 space straight S space mol space straight L to the power of negative 1 end exponent Applying space the space law space of space solubility space product comma space space space straight K subscript sp space equals open square brackets Fe to the power of 3 plus end exponent close square brackets space open square brackets OH to the power of minus close square brackets cubed space equals space straight S space cross times space left parenthesis 3 straight S right parenthesis cubed space equals space 27 straight S to the power of 4
therefore space space space straight S equals space open parentheses straight K subscript sp over 27 close parentheses to the power of 1 divided by 4 end exponent space equals space open parentheses fraction numerator 1 space cross times space 10 to the power of negative 38 end exponent over denominator 27 end fraction close parentheses to the power of 1 divided by 4 end exponent space space space space space space equals space open parentheses 100 over 27 cross times space 10 to the power of negative 40 end exponent close parentheses to the power of 1 divided by 4 end exponent space space space space space space equals left parenthesis 3.70 space cross times space 10 to the power of negative 40 end exponent right parenthesis to the power of 1 divided by 4 end exponent space space space space space space equals 1.387 space cross times space 10 to the power of negative 10 end exponent space mol space straight L to the power of negative 1 end exponent
therefore space space Molarity space of space Fe to the power of 3 plus end exponent space ions comma space space space space space space open square brackets Fe to the power of 3 plus end exponent close square brackets space equals space straight S space equals space 1.387 space cross times space 10 to the power of negative 10 end exponent straight M Molarity space of space OH to the power of minus space ions comma space space space space space space space open square brackets OH to the power of minus close square brackets space equals space 3 straight S space equals space 4.16 space cross times space 10 to the power of negative 10 end exponent straight M

Some More Questions From Equilibrium Chapter

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?

What is the role of forward and backward reactions at equilibrium of a reversible reaction?

What is a forward reaction?

What condition favours a reversible reaction?

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