Solubility of Th_3(PO_4)_4 (M_w = m) is w g | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The dissociation of $Th_3(PO_4)_4$ is: $Th_3(PO_4)_4 \rightleftharpoons 3Th^{4+} + 4PO_4^{3-}$.Let solubility be $s \ mol/L$.Given solubility $= w

Vedclass · Accepted Answer

$6.912 	imes 10^{10} \left( \frac{w}{m} ight)^7$

Vedclass · Answer

(A) The dissociation of $Th_3(PO_4)_4$ is: $Th_3(PO_4)_4 ightleftharpoons 3Th^{4+} + 4PO_4^{3-}$.Let solubility be $s \ mol/L$.Given solubility $= w \ g / 100 \ mL = 10w \ g / L = \frac{10w}{m} \ mol/L$.So,$s = \frac{10w}{m}$.$K_{sp} = [Th^{4+}]^3 [PO_4^{3-}]^4 = (3s)^3 (4s)^4$.$K_{sp} = 27s^3 	imes 256s^4 = 6912s^7$.Substituting $s = \frac{10w}{m}$:$K_{sp} = 6912 	imes \left( \frac{10w}{m} ight)^7 = 6912 	imes 10^7 	imes \left( \frac{w}{m} ight)^7$.$K_{sp} = 6.912 	imes 10^{10} \left( \frac{w}{m} ight)^7$.

Solubility of $Th_3(PO_4)_4$ $(M_w = m)$ is $w \ g / 100 \ mL$ at $25 \ ^oC$. The $K_{sp}$ is:

Similar Questions

If $S$ and $K_{sp}$ are respectively solubility and solubility product of a sparingly soluble binary electrolyte,then

Solubility of a binary sparingly soluble salt is $1.12 \times 10^{-4} \ g \ dm^{-3}$. Calculate its solubility product $(K_{sp})$ if the molar mass of the salt is $112 \ g \ mol^{-1}$.

The relationship between solubility $(S)$ and solubility product $(K_{sp})$ for a salt of type $A_x B_y$ is given by:

The solubility product of $Mg(OH)_2$ is $1.8 \times 10^{-11}$ at $298 \ K$. What is its solubility in $mol \ dm^{-3}$?

For a sparingly soluble salt $A_pB_q$,the relationship of its solubility product $(L_S)$ with its solubility $(S)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module