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048 mol0405 l 012 m c2h3o2 01104 mol0405 l 0273

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Unformatted text preview: 0.1104 mol/0.405 L = 0.273 M HC2H3O2 + H2O init 0.12 M equil ~0.12 X H3O+ + C2H3O2~10-7 M x x = 7.52 x 10-6 M = [H3O+] 5. 0.273 M ~0.273 Y pH = 5.12 Calculate the pH of 10-5 M HCl. This is a solution of a strong acid and [H3O+] ~ [acid]. [H3O+] = 10-5 M Y pH = 5.0 Chem 102 D. Miller Solutions to Review Problems for Acid/Base Chemistry 6. If 0.050 mL of 6.0 M HCl is added to 400 mL of 10-5 M HCl, what is the resulting pH? The added HCl is combines with the initial HCl to form a new strong acid solution. mol H3O+ added = (6.0 mol/L)(0.000050 L) = 0.00030 mol init mol H3O+ = (10-5 mol/L)(0.400 L) = 0.000004 mol total mol H3O+ = 0.000304 mol [H3O+] = 0.000304 mol/0.40005 L = 7.6 x 10-4 M pH = 3.12 (Notice that the pH of this unbuffered solution decreased by almost 2 pH units with this small addition of acid . Adding 100 times this volume of acid to a buffer (Prob. 4) changed the buffer pH by only 0.34 unit.) C...
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