The pH of a solution prepared by mixing 74.9 mL of 0.125 M Mg(OH)2 and 214.4...

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Answer 1

Answer #1

Volume of Mg(OH)2 = 74.9 ml

Concentration of Mg(OH)2 = 0.125 M

Moles of Mg(OH)2 = 74.9 x 0.125 / 1000 = 0.0093625‬ moles

Moles of OH- ions = 0.0093625‬ moles x 2 = 0.018725 Moles

Because one mole of Mg(OH)2 will be having 2 moles of OH- ions

Volume of HCl = 214.4 ml

Concentration of HCl = 0.125 M

Moles of HCl = 214.4 x 0.125 /1000 = 0.0268 Moles

HCl solution is in excess

Moles of HCl after reaction = 0.018725 Mol - 0.0268 Mol = 0.008075‬ moles

Volume of total solution = 74.9 ml + 214.4 ml = 289.3 ml

Concentration of H+ ions = 0.008075‬ x 1000 /289.3 = 0.0279 M

pH = - Log H+

substutute the H+ ion concentration and find out pH

pH = -Log 0.0279

pH = 1.55

Hence the pH of the final solution is 1.55

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Answer 2

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