Solution 0.1 M HCI 1.0 M HC1 0.1 M CH3COOH 1.0 M CH3COOH 0.1 M NaOH 0.1 M NH3 pH = -log [H0] = [H;O] + [H₂01 <= ?? (x1)

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Solution 0 1 M Hci 1 0 M Hc1 0 1 M Ch3cooh 1 0 M Ch3cooh 0 1 M Naoh 0 1 M Nh3 Ph Log H0 H O H 01 X1 1 (68.01 KiB) Viewed 10 times

Solution 0.1 M HCI 1.0 M HC1 0.1 M CH3COOH 1.0 M CH3COOH 0.1 M NaOH 0.1 M NH3 pH = -log [H0] = [H;O] + [H₂01 <= ?? (x1) Initial Changes Equilibrinm Ka-> pH paper = 2 1 3 2 13 9 3. Calculate Ka for 0.1 M acetic acid from the pH obtained from pH meter in Part I. antilog(-pH)= x][x] [0.1-x] = -X 0.1 0.1-x [H]+ CH3COOH +H₂ 2 pH/pH Range pH meter 2.46 1.38 2.91 2.62 11.97 8.74 I Ell 140 +x Methyl Red 3 <4 <4 4-6 <4 2 0 +x 9 Methyl Orange 1-2 0-1 2-4 1-3 (4 marks)