Enter the solubility-product expression for Al(OH)3(s). [Al3+][OH−]3

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Enter the solubility-product expression for Al(OH)3(s).  [Al3+][OH−]3

Daniel Cormack · Accepted Answer

Step 1  S be the solubility of Al(OH)3  Ksp=[Al3+][OH−]3=(S)(3S)3=27S4  S4=Ksp27=27×10−1127×10=1×10−12  S=1×10−3mil L−1  Step 2  Solution of Al(OH)3: Molar mass of Al(OH)3 78 g. Therefore, solubility of Al(OH)3 in  gL−1=1×103×78gL−1=78×10−3gL−1=7.8×10−2gL−1  Step 3  pH of the solution: S=1×10−3mol L−1  [OH]=3S=3×1×10−3=3×10−3  p[OH]=3−log⁡3  pH=14−pOH=11+log⁡3=11.4771

vrangett · Accepted Answer

Step 1 Suppose the solubility is S mol L−1. Then Al(OH)3⇒Al3++3OH−S3S Ksp=S×(3s)3=27S4 ∴27S4=2.7×10−11 or S4=10−12 or S=10−3mol L−1 Molar mass of Al(OH)3 in gL−1=10−3×78=7.8×10−2gL−1 [OH−]=3S=3×10−3mol L−1 ∴pOH=−log⁡(3×10−3=3−0.4771=3.5229 pH=14−2.5229=11.4771

Don Sumner · Accepted Answer

Step 1 The equation is as follows: Al(OH)3(s) AL3+(aq)+3OH−(aq) The solubility equation of Al(OH)3(s) is given above. The ionization of AL(OH)3(s) gives Al3+ions and OH−ions in water. The coefficient of Al3+ ion is one whereas the coefficient of OH− ion is three. Step 2 The solubility product expression Ksp is as follows: Ksp=[Al3+][OH−]3 Solubility-product expression for Al(OH)3(s) is Ksp=[Al3+][OH−]3 NSK Ksp of Al(OH)3(s) is the product of molar solubility of ions Al3+ and OH− formed in the ionization of Al(OH)3(s). Therefore, the Ksp expression for Al(OH)3(s) is Ksp=[Al3+][OH−]3 NSK Solubility-product expression for Al(OH)3(s) is Ksp=[Al3+][OH−]3

Enter the solubility-product expression for Al(OH)3(s). [Al^{3+}][O

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