g&g_ch_2_p_1-6,_11

g&g_ch_2_p_1-6,_11 - G&G Ch. 2 Problem 1 (pg. 49)...

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Problem 1 (pg. 49) a. 5x10 -4 M HCl All acids and bases in this problem can be assumed to proceed to complete dissociation. And remember, pH = log 10 [H 3 O + ] and pOH = log 10 [OH - ] pH = log 10 (5x10 -4 ) = 3.3 b. 7x10 -5 M NaOH pOH = log 10 (7x10 -5 ) = 4.2 pH = 14 – pOH = 9.8 c. 2μM HCl pH = log 10 (2x10 -6 ) = 5.7 d. 3x10 -2 M KOH pOH = log 10 (3x10 -2 ) = 1.5 pH = 14 – pOH = 12.5 e. 0.04mM HCl pH = log 10 (4x10 -5 ) = 4.4 f. 6x10 -9 M HCl pH = log 10 (1.06x10 -7 ) = 6.97 Because the HCl added to water in this instance is so dilute, any adjustment made to the pH by such a small amount of added H 3 O + is only noticeable if it is added to the contribution of H 3 O + from the self- hydrolysis of H 2 O (1x10 -7 M). Normally, our acids and bases are concentrated enough to far outweigh the contribution from self-hydrolysis (compare the exponents), which is why we usually leave it out. [HCl] total = 6x10 -9 M + 1x10 -7 M = 1.06x10 -7 M 2μM = 2x10 -6 M 0.04mM = 4x10 -5 M
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Problem 2 (pg. 49) a. [H + ] in vinegar pH = 2.9 1.26mM b. [H + ] in saliva pH = 6.6 250nM c. [H + ] in household ammonia Use Table 2.3 (pg. 40) for pH values. [H 3 O + ] = 10 -2.9 = 1.26x10 -3 M = [H 3 O + ] = 10 -6.6 = 2.5x10 -7 M = pH = 11.4 4pM [H 3 O + ] = 10 -11.4 = 4x10 -12 M = d. [OH - ] in milk of magnesia pH = 10.3 200μM [OH - ] = 10 -3.7 = 2x10 -4 M = pOH = 3.7 e. [OH - ] in beer pH = 4.5 320pM [OH - ] = 10 -9.5 = 3.2x10 -10 M = pOH = 9.5 f. [H + ] inside a liver cell pH = 6.9 130nM [H 3 O + ] = 10 -6.9 = 1.3x10 -7 M = I have put my answers in commonly used unit prefixes. Those in G&G vary in form. Here’s a little practice for unit conversions: convert all the answers in G&G to M, and
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This note was uploaded on 07/16/2008 for the course BIS 102 taught by Professor Hilt during the Summer '08 term at UC Davis.

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g&g_ch_2_p_1-6,_11 - G&G Ch. 2 Problem 1 (pg. 49)...

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