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12.5mL sample of 0.098 M Ba(OH)2 is titrated with 0.111M HCl solution. (Questions all go together)...

12.5mL sample of 0.098 M Ba(OH)2 is titrated with 0.111M HCl solution.

(Questions all go together)

1. Calculate pH of Ba(OH)2 solution.

2. Calculate pH after 3.44mL of HCI have been added.

3. What is the pH at equivalence point?

4. What is pH after adding 3.44mL of HCI beyond equivalence point?

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Answer #1

17 [Ba(OH)₂ 0.098M Ba(OH)2 → Ba 2+ 20H concentration - flow] 2x 0.098 M 0.196M Рон - - log [ori -log (0.196M) pot - 0.71 pH =[OH 0.065 M x2 0.130M роҢ : -log (0.130) POH = 0.89 PH 14-0-89 PH 13.11 (3) At equivalence point : 2x Moles of Ha moles of BaMoles of HCl O.ILIM X 25.54ml = 2.835 Millimales Moles of Balora 1.225 mmeles I mele BalOH)2 neutralize by 2 meles of Hee. 1.

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