If the salts $M_{2} X$ , $Q Y_{2}$ and $P Z_{3}$ have same solubilities (<<<1) ,their $K_{s p}$ values are related as:

M_{2} X = Q Y_{2} = P Z_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

M_{2} X = Q Y_{2} < P Z_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

M_{2} X = Q Y_{2} > P Z_{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

M_{2} X < Q Y_{2} = P Z_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $M_{2} X = Q Y_{2} > P Z_{3}$
Let S be the solubility in moles/ litre.
The expression for the solubility product of $M_{2} X$ is $(2 S)^{2} S = 4 S^{3}$
The expression for the solubility product of $Q Y_{2}$ is $S (2 S)^{2} = 4 S^{3}$
The expression for the solubility product of $P Z_{3}$ is $S (3 S)^{3} = 27 S^{4}$
Since $S <<< 1$ , $M_{2} X = Q Y_{2} > P Z_{3}$ .

Suggest Corrections

Similar questions

M_{2} X, Q Y_{2}

P Z_{3}

K_{s p}

M_{2} X

Q Y_{2}

P Z_{3}

K_{s p}

p x^{2} - q y^{2} = 0

∣ ∣ ∣ ∣ ∣ \begin{matrix} x & x^{2} & 1 + {p x}^{3} y & y^{2} & 1 + {p y}^{3} z & z^{2} & 1 + {p z}^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣

= (1 + p x y z) (x - y) (y - z) (z - x)

M_{2} X, M X

M X_{3}

Join BYJU'S Learning Program