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Question

The first ionization constant of H2S is 9.1×108. Calculate the concentration of HS ion in its 0.1 M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H2S is 1.2×1013, calculate the concentration of S2 under both conditions.

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Solution

(i) To calculate [HS] in absence of HCl:
Let, [HS]=x M.
H2SH++HS
The initial concentrations of H2S,H+ and HS are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-x M, x M and x M respectively.
Ka=[H+][HS][H2S]
9.1×108=x×x0.1x
In the denominator, 0.1-x can be approximated to 0.1 as x is very small.
9.1×108=x×x0.1
x=9.54×105M=[HS]
(ii) To calculate [HS] in presence of HCl:
Let [HS]=y M.
H2SH++HS
The initial concentrations of H2S,H+ and HS are 0.1 M, 0 M and 0 M respectively.
Their final concentrations are 0.1-y M, y M and y M respectively.
Also, HClH++Cl
For HCl [H+]=[Cl]=0.1M
Ka=[HS][H+][H2S]
Ka=y(0.1+y)0.1y
In the denominator, 0.1-y can be approximated to 0.1 and in the numerator, 0.1+y can be approximated to 0.1 as y is very small.
9.1×108=y×0.10.1
y=9.1×108M=[HS]

(iii) To calculate [S2] in absence of 0.1 M HCl:
HSH++S2
[HS]=9.54×105M
Let [S2]=X M.
[H+]=9.54×105 M
Ka=[H+][S2][HS]
1.2×1013=9.54×105×X9.54×105
X=1.2×1013M=[S2]

(iv) To calculate [S2] in presence of 0.1 M HCl:
Let [S2]=X M.
[HS]=9.1×108 M
[H+]=0.1 M (from HCl)
Ka=[H+][S2][HS]
1.2×1013=0.1×X9.1×108
X=1.092×1019=[S2]

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