An amount of solid Ba(OH) 2 isdissolved in water to make a final solution of 146

Question

An amount of solid Ba(OH)2 isdissolved in water to make a final solution of 1460 mL.
If the pH of this solution is found to be 12.39, how manygrams of the base were added to make this solution ? An amount of solid Ba(OH)2 isdissolved in water to make a final solution of 1460 mL.
If the pH of this solution is found to be 12.39, how manygrams of the base were added to make this solution ?

Explanation / Answer

pH = 12.39=> pOH = 14 - pH = 1.61                => [OH-] = 10-pOH = 0.0245M Since Ba(OH)2 is strong base, it dissociatescompletely.                 Ba(OH)2 --> Ba2+ +2OH-   Notice that [OH-]/2 =[Ba(OH)2] =>[Ba(OH)2] = 0.0245M / 2 = 0.01225M Mole of Ba(OH)2 needed: n = 0.01225M * 1460mL *1L/1000mL = 0.0179(mol) Mass of Ba(OH)2 needed to add:        m = Mn = 171.2g/mol *0.0179mol = 3.07g

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