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По закон препоръчителен легитимен ioners opløselighed стръмен подплата тържище

$\begin{array}{llllll} \textup{opl\o selighedsprodukt:}\\&& \left [ Ba^{2+} \right ] \cdot \left [ S{O_4}^{2-} \right ]=1.5\cdot 10^{-9}\;\mathrm{M^2}\\\\ \textup{n\aa r der i forvejen er}\\ 0.005\;\mathrm{M}\; Ba^{2+}\textup{:}\\&&(0.005+x)\cdot x=1.5\cdot 10^{-9}\\\\&&x^2+0.005x-1.5\cdot 10^{-9}=0,\quad x>0\\\\&& x=3.0\cdot 10^{-7}\;\mathrm{\frac{mol}{L}}=7.0\cdot 10^{-5}\;\mathrm{\frac{g}{L}}=0.07\;\mathrm{\frac{mg}{L}} \\\\ \textup{I en 0.005 M opl\o sning}\\\textup{af }Ba^{2+}\textup{-ioner}\\ \textup{kan der opl\o ses}&&0.07\;\mathrm{mg }\; BaSO_4\textup{ pr liter opl\o sning.} \end{array}$

$\begin{array}{lllll} &\left [ Al^{3+} \right ]\cdot \left [ F^- \right ]^3=K_{sp}\\\\\textup{kaldes den mol\ae re}\\\textup{opl\o selighed for }s\\\textup{haves:}&s\cdot (3s)^3=1.10\cdot 10^{-3}\; M^4\\\\&27s^4=1.10\cdot 10^{-3}\; M^4\\\\&s=\left ( \frac{1.10\cdot 10^{-3}\; M^4}{27} \right ) ^{\frac{1}{4}}=0.079893\; M\\\textup{omregnet}\\\textup{opl\o selighed}&\left (0.079893\; \frac{mol}{L} \right )\cdot \left ( 83.977\; \frac{g}{mol} \right )=6.71\; \frac{g}{L}=0.671\; \frac{g}{100\; mL} \end{array}$