Buffering against pH changes in biologicalsystems. A compound has a pKa of 7.4.

Question

Buffering against pH changes in biologicalsystems. A compound has a pKa of 7.4. To 100 mL of a 1.0 Msolution of this compound at pH 8.0 is added 30 mL of 1.0 Mhydrochloric acid. The resulting solution is pH: (a.) 6.5 (b.) 6.8 (c.) 7.2 (d.) 7.4 (e.) 7.5 (In order to receive credit, all calculations that are used tolead to the answer must be shown. It should be obvious whatassumptions and mathematical relationships were used to arrive atthe answer.) A compound has a pKa of 7.4. To 100 mL of a 1.0 Msolution of this compound at pH 8.0 is added 30 mL of 1.0 Mhydrochloric acid. The resulting solution is pH: (a.) 6.5 (b.) 6.8 (c.) 7.2 (d.) 7.4 (e.) 7.5 (In order to receive credit, all calculations that are used tolead to the answer must be shown. It should be obvious whatassumptions and mathematical relationships were used to arrive atthe answer.)

Explanation / Answer

pH = pKa + log ([A-]/[HA]) 8 = 7.4 + log ([A-]/[HA]) 0.6 = log ([A-]/[HA]) ([A-]/[HA]) = 3.98/1 Volume of A- = 3.98/4.98 x 100 = 79.92mL c(A-) = n/v n(A-) = cv = 1 x 0.07992 = 0.07992mol Volume of HA = 1/4.98 x 100 = 20.08mL c(HA) = n/v n(HA) = cv = 1 x 0.02008 = 0.02008mol n(HCl) = cv = 1 x 0.03 = 0.03mol New n(A-) = 0.07992 - 0.03 = 0.04992mol New n(HA) = 0.02008 + 0.03 = 0.05008mol Total volume = 100 + 30 = 130 mL c(A-) = n/v = 0.04992/0.13 = 0.384M c(HA) = n/v = 0.05008/0.13 = 0.3852M pH = pKa + log ([A-]/[HA]) pH = 7.4 + log (0.384/0.3852) pH = 7.40 The answer is (d) Hope this helps!

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