Solution 0.1 MHCI 1.0 MHCI 0.1 M CH3COOH 10M CHCOOH 0.1 M NaOH 0.1 MNHS pH paper pH = -log [HO]+= 5.15 = [H₂O]+=(x) Tabl

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answerhappygod
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Solution 0.1 MHCI 1.0 MHCI 0.1 M CH3COOH 10M CHCOOH 0.1 M NaOH 0.1 MNHS pH paper pH = -log [HO]+= 5.15 = [H₂O]+=(x) Tabl

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Solution 0 1 Mhci 1 0 Mhci 0 1 M Ch3cooh 10m Chcooh 0 1 M Naoh 0 1 Mnhs Ph Paper Ph Log Ho 5 15 H O X Tabl 1
Solution 0 1 Mhci 1 0 Mhci 0 1 M Ch3cooh 10m Chcooh 0 1 M Naoh 0 1 Mnhs Ph Paper Ph Log Ho 5 15 H O X Tabl 1 (43.13 KiB) Viewed 9 times
Solution 0.1 MHCI 1.0 MHCI 0.1 M CH3COOH 10M CHCOOH 0.1 M NaOH 0.1 MNHS pH paper pH = -log [HO]+= 5.15 = [H₂O]+=(x) Table Ka=? 2 1 3 2 13 9 pH meter 2.46 1.38 2.91 2.62 pH/pH Range 11.97 8.74 4. Calculate the Ka for 0.1 M acetic acid from the pH obtained for Solution Y in Part II Methyl Red <4 <4 4-6 <4 Methyl Orange 1-2 0-1 2-4 1-3 (3 marks) (Ctrl)
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