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Unformatted text preview: @pH 10.2 NH 2 = 0.8 eqvivalents to get to pH 8.6,which is pk a 1 and therefore 10% titrated (0.1 equivalent), we have to add 0.7 equivalents of HCL 1 equiv. = 0.1mol (0.1Mx1.0L) 0.7 equiv. = 0.07mol 5M = ? . ? ? x = 14ml d) When 99% of the glycine is in its -NH 3 + form, what is the numerical relation between the pH of the solution and the pK a of the amino group? pH = pk a + log ~ pH = pk a + log ?? pH = pk a 2 result: at 1% titrated pH = pk a 2 at 99% titrated pH = pk a + 2...
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- Winter '09