Question

3. A solution is 0.100 M in H2S. Which ion HS-or S2-
will be present in larger concentrations?​

Answer 1

Answer:

$H </p><p>2</p><p> </p><p> S⇌H </p><p>+</p><p> +HS </p><p>−</p><p> </p><p>K </p><p>a </p><p>1</p><p> </p><p> </p><p> </p><p> = </p><p>[H </p><p>2</p><p> </p><p> S]</p><p>[H </p><p>+</p><p> ][HS </p><p>−</p><p> ]</p><p> </p><p> </p><p>Further HS </p><p>−</p><p> ⇌H </p><p>+</p><p> +S </p><p>2−</p><p> </p><p>K </p><p>a </p><p>2</p><p> </p><p> </p><p> </p><p> = </p><p>[HS </p><p>−</p><p> ]</p><p>[H </p><p>+</p><p> ][S </p><p>2−</p><p> ]</p><p> </p><p> </p><p>Multiplying both the equations</p><p>K </p><p>a </p><p>1</p><p> </p><p> </p><p> </p><p> ×K </p><p>a </p><p>2</p><p> </p><p> </p><p> </p><p> = </p><p>[H </p><p>2</p><p> </p><p> S]</p><p>[H </p><p>+</p><p> ] </p><p>2</p><p> [S </p><p>2−</p><p> ]</p><p> </p><p>$

Multiplying both the equations

$K </p><p>a </p><p>1</p><p> </p><p> </p><p> </p><p> ×K </p><p>a </p><p>2</p><p> </p><p> </p><p> </p><p> = </p><p>[H </p><p>2</p><p> </p><p> S]</p><p>[H </p><p>+</p><p> ] </p><p>2</p><p> [S </p><p>2−</p><p> ]</p><p> </p><p>$

Due to common ion, the ionisation of H 2S is suppressed and the [H+] in solution is due to the presence of 0.3M HCl

$[S </p><p>2−</p><p> ]= </p><p>[H </p><p>+</p><p> ] </p><p>2</p><p> </p><p>K </p><p>a </p><p>1</p><p> </p><p> </p><p> </p><p> ×K </p><p>a </p><p>2</p><p> </p><p> </p><p> </p><p> [H </p><p>2</p><p> </p><p> S]</p><p> </p><p> =1.44×10 </p><p>−20</p><p> M$

Putting the value of [S2−] in eq. (ii)

$1.3×10 </p><p>−13</p><p> = </p><p>[HS </p><p>−</p><p> ]</p><p>0.3×1.44×10 </p><p>−20</p><p> </p><p> </p><p> </p><p>or [HS </p><p>−</p><p> ]=3.3×10 </p><p>−8$