Question

The first ionization constant of H_2S is 9.1 × 10^–8. Calculate the concentration of HS^– ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H_2S is 1.2 × 10^–13, calculate the concentration of S^2– under both conditions.
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Answer 1

Answer

Explanation:

1)To calculate [HS − ] in absence of HCl:

Let, [HS − ]=x M.

H 2 S⇌H + +HS −

The initial concentrations of

H 2 S,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.

K a =

$\{ \[H+][HS−]</p><p>}{h2s}$

$9.1×10−8 \\ \\ = \{x \timesx}{0.1} </p><p></p><p></p><p>$

$x=9.54×10 { - }^{5} \\ M=[HS−]</p><p></p><p>$

2)To calculate [HS − ] in presence of HCl:

Let [HS − ]=y M.

H 2 S⇌H + +HS −

The initial concentrations of H 2 S,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, HCl⇌H + +Cl −

For HCl [H + ]=[Cl − ]=0.1M

K a =

$= \{hs - h + }{h2s}$

K a =

$\{y(0.1+y)}{0.1 - y} </p><p></p><p></p><p>$

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.

$9.1×10 { - }^{8} \\ = {y\times 0.1}{0.1} </p><p></p><p>$

$y=9.1×10 { - }^{8} M=[HS−]</p><p></p><p>$

3)To calculate [S 2− ] in absence of 0.1 M HCl:

$HS − ⇌H +S {2}^{ - }$

$[HS − ]=9.54×10 { - }^{5} M$

Let [S 2− ]=X M.

$[H + ]=9.54×10 { - }^{5} M$

K a =

$= {[H+][S2−]}{hs - } </p><p>$

$1.2 \times 10 { - }^{13 } =$

${9.4 \times 10 { - }^{5} \times x}{9.54 \times 10 { - }^{5} }$

$X=1.2×10 { - }^{13} M=S { - }^5$

4) To calculate [S 2− ] in presence of 0.1 M HCl:

$Let [S </p><p> {2}^{ - } </p><p> ]=X </p><p>′</p><p> M.$

$[HS−]=9.1×10 { - }^{8} M$

[H + ]=0.1 M (from HCl)

ka =

$= {[H+][S {2}^{ - } ]}{hs - } </strong></p><p><strong>[tex] = \{[H+][S {2}^{ - } ]}{hs - }$

$1.2×10 { - }^{13} =$

$\{0.1×X′</strong></p><p><strong>[tex] \{0.1×X′}{9.1 \times 10 { - }^{8} }$

$X′=1.092×10 { - }^{19} =[S {2}^{ - } ]$

Answer 2

Answer:

Concentration of HS⁻ ions when HCl is absent = 9.539 × 10⁻⁵ M.

Concentration of HS⁻ ions when HCl is present = 9.1 × 10⁻⁸ M.

Concentration of S²⁻ when HCl is absent = 1.2 × 10⁻¹³ M.

Concentration of S²⁻ when HCl is present = 1.092 × 10⁻¹⁹M.

Explanation:

Given:

• The first ionization constant of H₂S = 9.1 × 10⁻⁸
• Concentration of H₂S = 0.1 M
• Concentration of HCl = 0.1 M
• Second dissociation constant of H₂S = 1.2 × 10⁻¹³

To Find:

• Concentration of HS⁻ ions and S²⁻ ions when HCl is present and when HCl is absent

Solution:

a) Concentration of HS⁻ ions:

i) In the absence of HCl:

Given first dissociation constant K = 9.1 × 10⁻⁸

                    $\tt H_2S \rightleftharpoons H^++HS^-$

Initial conc:  0.1       0      0

Final conc:  0.1 - x    x      x

That is,

$\tt K =\dfrac{[H^+][HS^-]}{[H_2S]}$

$\implies \tt \dfrac{x\times x}{(0.1-x)}$

Here x ≈ 0, hence,

$\tt \dfrac{x^2}{0.1} =9.1\times 10^{-8}$

$\tt x^2=9.1\times 10^{-8}\times 0.1$

$\tt x=\sqrt{9.1\times 10^{-9}}$

$\implies \tt 9.539\times 10^{-5}\:M$

Hence concentration of HS⁻ ions when HCl is absent is 9.539 × 10⁻⁵ M.

ii) In the presence of HCl,

                   $\tt H_2S \rightleftharpoons H^++HS^-$

Initial conc:  0.1       0      0

Final conc:  0.1 - y    y      y

Also,

           $\tt HCl \rightleftharpoons H^++Cl^-$

Conc:  0.1       0.1    0.1

$\tt K =\dfrac{[H^+][HS^-]}{[H_2S]}$

$\tt \implies \dfrac{y\times (0.1+y)}{0.1-y}$

y ≈ 0, therefore,

$\tt \dfrac{y\times 0.1}{0.1}=9.1\times 10^{-8}$

$\tt y=9.1\times 10^{-8}\:M$

Hence concentration of HS⁻ ions when HCl is present is 9.1 × 10⁻⁸ M.

b) Concentration of S²⁻ ions,

i) When HCl is absent:

Given that second dissociation constant is 1.2 × 10⁻¹³

                         $\tt HS^{-} \rightleftharpoons H^++S^{2-}$

Conc: 9.539 × 10⁻⁵   9.539 × 10⁻⁵    z

$\tt K=\dfrac{[H^+][S^{2-}]}{[HS^-]}$

$\implies \tt \dfrac{9.539\times 10^{-5}\times z}{9.539\times 10^{-5}}$

$\tt z=1.2\times 10^{-13}\: M$

Hence concentration of S²⁻ when HCl is absent is 1.2 × 10⁻¹³ M.

ii) When HCl is present:

$\tt K=\dfrac{[H^+][S^{2-}]}{[HS^-]}$

$\tt K = \dfrac{0.1\times [S^{2-}]}{9.1\times 10^{-8}} =1.2\times 10^{-13}$

$\implies \tt [S^{2-}]=\dfrac{1.2\times 10^{-13}\times 9.1\times 10^{-8}}{0.1}$

$\implies \tt 1.092\times 10^{-19}\: M$

Hence the concentration of S²⁻ when HCl is present is 1.092 × 10⁻¹⁹M.