CHM 116 Lecture 17: Chapter 17 Addition Aspects of Equilibrium
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Answer: 2 because hno3 is a strong acid, so it cannot form a buffer solution. An effective buffer is within 1 ph unit of the pka. For example, lactic acid has a ka of 1. 4 x 10-4. As we saw lactic acid has a pka of 3. 85. 0. 12 m lactic acid and 0. 10 m sodium lactate. 0. 52 m lactic acid and 0. 5 m sodium lactate. 1. 03 m lactic acid and 0. 01 m sodium lactate. You typically want 0. 1 m and greater for the buffer (in this case sodium lactate) ph of a buffered solution. Ka = 1. 4 x 10-4 = [h+][c3h5o3-] / [hc3h5o3] = (x(0. 1 + x)) / (0. 12 - x) x = 1. 7 x 10-4 therefore the ph of solution is: ph = -log[1. 7 x 10-4] = 3. 77. But here"s an easier equation than using an ice table. = -log ka = (-log[h+] - log([a-] / [ha]) pka = ph - log([a-] / [ha])
