Home
Class 12
CHEMISTRY
What are the concentration of H(3)O^(+)...

What are the concentration of `H_(3)O^(+) , HS^(+) and S^(2-) ` in `0..01` M `H_(2)S` solution , given ` {:(H_(2)S,+,H_(2)O,hArr,H_(3)O^(+),+,HS^(-),K_(at),=1.1 xx10^(-7),),(HS,+,H_(2)O,hArr,H_(3)O^(+),+,S^(2),K_(at),=1.0xx10^(-14),):}` .

Text Solution

AI Generated Solution

To find the concentrations of \( H_3O^+ \), \( HS^- \), and \( S^{2-} \) in a \( 0.01 \, M \) \( H_2S \) solution, we will use the given dissociation constants and the principles of equilibrium. ### Step 1: Write the dissociation reactions and expressions for \( K_a \) The dissociation of \( H_2S \) can be represented as follows: 1. For the first dissociation: \[ ...
Promotional Banner

Topper's Solved these Questions

  • IONIC EQUILIBRIUM

    FIITJEE|Exercise EXERCISE 5|1 Videos

  • IONIC EQUILIBRIUM

    FIITJEE|Exercise EXERCISE 6|1 Videos

  • IONIC EQUILIBRIUM

    FIITJEE|Exercise EXERCISE 3|1 Videos

  • HYDROCARBONS

    FIITJEE|Exercise SINGLE INTEGER ANSWER TYPE QUESTION|9 Videos

  • LIQUID SOLUTION

    FIITJEE|Exercise Single Integer Answer Type Question|10 Videos

Similar Questions

Explore conceptually related problems

The number of S^(2-) ions present in 1 L of 0.1 (M) H_(2)S solution having [H^(+)]=0.1 (M) is ( Given H_(2)ShArr2H^(+)+S^(2-) K_(a)=1.1xx10^(-21) )

What is [HS^(-)] & [S^(2-)] in solution of 0.01M HCl and 0.1 M H_(2)S ?(Given that for H_(2)S,K_(a_(1))=10^(-7)and K_(a_(2))=10^(-14))

2H_(2)O hArr H_(3)O^(o+) + overset(Theta)OH,K_(w) = 10^(-14) at 25^(@)C , hence K_(a) is

One "mole" of NH_(4)HS(s) was allowed to decompose in a 1-L container at 200^(@)C . It decomposes reversibly to NH_(3)(g) and H_(2)S(g). NH_(3)(g) further undergoes decomposition to form N_(2)(g) and H_(2)(g) . Finally, when equilibrium was set up, the ratio between the number of moles of NH_(3)(g) and H_(2)(g) was found to be 3 . NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g), K_(c)=8.91xx10^(-2) M^(2) 2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g), K_(c)=3xx10^(-4) M^(2) Answer the following: Assuming the volume due to solid NH_(4)HS is negligible what will be the density of the gaseous mixture in the above equilibrium system?

One "mole" of NH_(4)HS(s) was allowed to decompose in a 1-L container at 200^(@)C . It decomposes reversibly to NH_(3)(g) and H_(2)S(g). NH_(3)(g) further undergoes decomposition to form N_(2)(g) and H_(2)(g) . Finally, when equilibrium was set up, the ratio between the number of moles of NH_(3)(g) and H_(2)(g) was found to be 3 . NH_(4)HS(s) hArr NH_(3)(g)+H_(2)S(g), K_(c)=8.91xx10^(-2) M^(2) 2NH_(3)(g) hArr N_(2)(g)+3H_(2)(g), K_(c)=3xx10^(-4) M^(2) Answer the following: To attain equilibrium, how much % by weight of folid NH_(4)HS got dissociated?