Consider two Group IV metal ions $\mathrm{X}^{2+}$ and $\mathrm{Y}^{2+}$. A solution containing $0.01 \mathrm{MX}^{2+}$ and $0.01 \mathrm{MY}^{2+}$ is saturated with $\mathrm{H}_{2} \mathrm{~S}$. The pH at which the metal sulphide YS will form as a precipitate is ______ (Nearest integer) (Given : $\mathrm{K}_{\mathrm{sp}}(\mathrm{XS})=1 \times 10^{-22}$ at $25^{\circ} \mathrm{C}$, $\mathrm{K}_{\mathrm{sp}}(\mathrm{YS})=4 \times 10^{-16}$ at $25^{\circ} \mathrm{C}$, $\left[\mathrm{H}_{2} \mathrm{~S}\right]=0.1 \mathrm{M}$ in solution, $\mathrm{K}_{\mathrm{a} 1} \times \mathrm{K}_{\mathrm{a} 2}\left(\mathrm{H}_{2} \mathrm{~S}\right)=1.0 \times 10^{-21}$, $\log 2=0.30, \log 3=0.48, \log 5=0.70$ )

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Equilibrium

Consider two Group IV metal ions

X^{2 +}

and

Y^{2 +}

.

A solution containing

0.01 MX^{2 +}

and

0.01 MY^{2 +}

is saturated with

H_{2} S

. The pH at which the metal sulphide YS will form as a precipitate is ______
(Nearest integer)
(Given :

K_{sp} (XS) = 1 \times 1 0^{- 22}

2 5^{\circ} C

K_{sp} (YS) = 4 \times 1 0^{- 16}

2 5^{\circ} C

[H_{2} S] = 0.1 M

in solution,

K_{a 1} \times K_{a 2} (H_{2} S) = 1.0 \times 1 0^{- 21}

lo g 2 = 0.30, lo g 3 = 0.48, lo g 5 = 0.70

)

Answer: 4

Precipitation of

YS

starts when

[S^{2 -}] = \frac{K _{s p} ( YS )}{[ Y ^{2 +} ]} = \frac{4 \times 1 0 ^{- 16}}{0.01} = 4 \times 1 0^{- 14} M

. The sulphide level is set by

[S^{2 -}] = \frac{K _{a 1} K _{a 2} [ H _{2} S ]}{[ H ^{+} ] ^{2}} = \frac{1 0 ^{- 21} \times 0.1}{[ H ^{+} ] ^{2}} = \frac{1 0 ^{- 22}}{[ H ^{+} ] ^{2}}

. Equating the two gives

[H^{+}]^{2} = \frac{1 0 ^{- 22}}{4 \times 1 0 ^{- 14}} = 2.5 \times 1 0^{- 9}

, so

[H^{+}] = 5 \times 1 0^{- 5} M

and

pH = 5 - lo g 5 = 4.30

. To the nearest integer the pH is

4

JEE Main · 2026 · 24 Jan 2026 · Shift 1 · Integer answer