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The solubility products of three sparingly soluble salts $A B, A_2 B$ and $A B_3$ are respectively $4.0 \times 10^{-20}, 3.2 \times 10^{-11}$ and $2.7 \times 10^{-31}$.
The increasing order of their solubility is
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The increasing order of their solubility is
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Verified Answer
The correct answer is:
$A B < A B_3 < A_2 B$
The solubility product is defined as the product of concentrations of the ions raised to a power equal to the number of times the ions occur in the equation representing the dissociation of electrolyte at given temperature. (When the solution is saturated.)
(i) Electrolyte of the type $=A B$
$$
A B \leftrightarrow A^{+}+B^{-}
$$
If, $s=$ solubility.
$$
K_s=\left[A^{+}\right]\left[B^{-}\right]=4 \cdot 0 \times 10^{-20}=s^2
$$
(ii)
$$
\begin{aligned}
& A_2 B \leftrightarrow 2 A^{2+}+B^{2-} \\
& \quad K_s=\left[A^{+}\right]^2\left[B^{2-}\right]=3 \cdot 2 \times 10^{-11}=4 s^3
\end{aligned}
$$
(iii)
$$
\begin{aligned}
A B_3 \leftrightarrow A^{3+} & +3 B^{-} \\
K_s & =\left[A^{3+}\left[B^{2-}\right]^3=2 \cdot 7 \times 10^{-31}=27 s^4\right.
\end{aligned}
$$
The order of solubility is $A B < A B_3 < A_2 B$.
(i) Electrolyte of the type $=A B$
$$
A B \leftrightarrow A^{+}+B^{-}
$$
If, $s=$ solubility.
$$
K_s=\left[A^{+}\right]\left[B^{-}\right]=4 \cdot 0 \times 10^{-20}=s^2
$$
(ii)
$$
\begin{aligned}
& A_2 B \leftrightarrow 2 A^{2+}+B^{2-} \\
& \quad K_s=\left[A^{+}\right]^2\left[B^{2-}\right]=3 \cdot 2 \times 10^{-11}=4 s^3
\end{aligned}
$$
(iii)
$$
\begin{aligned}
A B_3 \leftrightarrow A^{3+} & +3 B^{-} \\
K_s & =\left[A^{3+}\left[B^{2-}\right]^3=2 \cdot 7 \times 10^{-31}=27 s^4\right.
\end{aligned}
$$
The order of solubility is $A B < A B_3 < A_2 B$.
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