2d:["$","li","1",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":2}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"H⁺ concentration is = 0.2 M."}}]]}] 38:T77d,H₂S

\rightleftharpoons

H⁺ + HS^

-

; K₁ = 1.0

\times

10^

-

72e:["$","li","2",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":3}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$38"}}]]}] 39:T94e,HS^

-

\rightleftharpoons

H⁺ + S^2

-

; K₂ = 1.2

\times

10^

-

132f:["$","li","3",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":4}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$39"}}]]}] 30:["$","li","4",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":5}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"H₂S

\\rightleftharpoons

S^2

-

+ 2H⁺"}}]]}] 3a:Ted3,K = K₁

\times

K₂ = 1.0

\times

10^

-

\times

1.2

\times

1.0^

-

13 = 1.2

\times

10^

-

2031:["$","li","5",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":6}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$3a"}}]]}] 32:["$","li","6",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":7}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"as K₁ and K₂ both are very low for this reaction so dissociation of H₂S and HS^

-

will be very low so, the produced H⁺ from this reaction will also be very low."}}]]}] 33:["$","li","7",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":8}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"So, we can say the concentration of H⁺ will be almost same as H⁺ in HCl."}}]]}] 34:["$","li","8",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":9}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"

\\therefore\\,\\,\\,

[ H⁺ ] = 0.2 M."}}]]}] 3b:T427,From the reaction, H₂S

\rightleftharpoons\,

2H⁺ + S^2

-

35:["$","li","9",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":10}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$3b"}}]]}] 36:["$","li","10",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":11}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"We get [ H⁺ ]² [ S^2

-

] = K

\\times

[ H₂ S ]"}}]]}] 3c:T102e,

\Rightarrow \,\,\,

[ S^2

-

] =

{{1.2 \times {{10}^{ - 20}} \times 0.1} \over {{{\left( {0.2} \right)}^2}}}

37:["$","li","11",{"className":"flex items-start gap-3","children":[["$","span",null,{"className":"mt-0.5 flex h-5 w-5 shrink-0 items-center justify-center rounded-md bg-surface-sunken font-mono text-[11px] text-ink-400","children":12}],["$","div",null,{"className":"min-w-0 max-w-[68ch] flex-1 text-[15px] leading-relaxed text-ink-600 [&_.katex-display]:my-3 [&_.katex-display]:overflow-x-auto","dangerouslySetInnerHTML":{"__html":"$3c"}}]]}]

More on Ionic Equilibrium

More on Ionic Equilibrium