Question

Solubility of thorium phosphate (molar mass$=m$) is $\frac{10w}{m}$ $\left(mol/litre\right)$ at ${25}^{o}C$.

The $K_{sp}$ of thorium phosphate is

• A. $6.912\times {10}^{10}\times {\left(\frac{w}{m}\right)}^7$
• B. $6.912\times {10}^{10}\times {\left(\frac{w}{m}\right)}^5$
• C. $6.912\times {10}^{10}\times {\left(\frac{w}{m}\right)}^4$
• D. $6.912\times {10}^{10}\times {\left(\frac{w}{m}\right)}^2$

Answer 1

The correct option is A $6.912\times {10}^{10}\times {\left(\frac{w}{m}\right)}^7$

${Th}_3{\left({PO}_4\right)}_4\rightleftharpoons 3\underset{3s}{{Th}^{4+}}+4\underset{4s}{{PO}_4^{3-}}$

$K_{sp}=3^3\cdot 4^4\cdot {\left(s\right)}^7=6912\times {\left(\frac{10w}{m}\right)}^7(s=\frac{10w}{m}mol/litre)$

$=6.912\times {10}^{10}\times {\left(\frac{w}{m}\right)}^7$