Question

Why does the A- initial not start with 10 nmols and why does HA initial not start at 0?

3c. (12 pts) The rxn below produces H. CH3CH2OH + NAD+ + CH3CHO + NADH + H+ Suppose the rxn is run in 1.0mL of 104 nM buffer whose pKa=8.50. If the rxn starts at pH=8.2, and the rxn releases 0.30 nmol of H", what will be the final pH? Hint: convert all numbers into nmoles. 8.2=8.5+log [A') gives [A") = 0.50 %A- - 0.50 x100 = 33.3% 1.50 1.00 %HA== xx100 = 66.7% 1.50 Convert to moles: 104 nmol 0.001L -= 10nmol IL Al-:10(0.33) = 3.3nmol HA:10(0.67) = 6.7nmol A1-(nmol) H+ (nmol) 3.3 0.3 C 0.3 -0.3 E 3.0 0 pH=8.5+log = 7.8+(-0.37) HA (nmol) 6.7 +0.3 7.0 pH = 8.13

Answer 1

1 Molar buffer does not mean [A​​​​​​- ] = 1 M . But it means [HA] + [A​​​​​​- ] = 1 M

In this problem also this is applicable and here the overall concentration is 10 nM .

Explained below

1 molar buffer means [ Acid] + [ sult] = 1.14 So, [HA] + [A-] = 1M. . 104nmole xl mL Hese, Overall concentrations 1000mL . = 10 nM i 8 .2 [A] . [HA] ::. So, [HA] + [A] = 10nn Now initial pH = 8.2 . se 8.2 = 85+log [HA] A 33.3%. G = 0.5 . HA = 66.7%. This indicate in the initial mixture of 10 mm [A-] = (0.33 X40)1M = 3.3 nm ( 33 % of 10nm) THA] = (0.67×10) hin= 6. Trim (67% of 10 nM). Then as shown in the ICE table pH = 8.5 + log i = 8.13 EM M .

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