Learning Goal: The following calculations are similar to those you will complete

Question

Learning Goal: The following calculations are similar to those you will complete as part of the lab Buffers work because the conjugate acid-base pair work together to neutralize the addition of H^+ or OH^- ions. Thus, for example, if H^+ ions are added to the acetate buffer described above, they will be largely removed from solution by the reaction of H^+ with the conjugate base: H^+ +CH_3COO^- rightarrow CH_3COOH Similarly, any added OH^- ions will be neutralized by a reaction with the conjugate acid: OH^- + CH_3COOH rightarrow CH_3COO^- + H_2O This buffer system is described by the Henderson-Hasselbalch equation P^H = PK_a + log [conjugate base]/[conjugate acid] The pK_a of acetic acid is 4.74. Your Buffer System: A beaker with 1.20 times 10^2 mL of an acetic acid buffer with a pH of 4.60 is sitting on a benchtop The total molarity of acid and conjugate base in this buffer is 0.100 M. How many moles of acid are in this buffer solution? 6.96 times 10^-3^moles Part B How many moles of the conjugate base are in this buffer solution? 5.04 times 10^-3^moles PartC If you add 5.90 mL of a 0.330 MHCl solution to the beaker, how much will the pH change? Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased. submit

Explanation / Answer

Part C

lets conjugate base = A-

conjugate acid = HA

pH = pKa + log(A- / HA)

4.6 = 4.74 + log(A-/HA)

log(A-/HA) = 4.6-4.74 = -0.14

[A- / HA] = 10-0.14 = 0.724 ----1

but he has given total molarity = 0.1

[A-] + [HA] = 0.1 -----2

using eq 1 and 2 find the individual concentrations

from 1 [A-] = 0.724 x [HA]

put this value in 2

0.724 [HA] + [HA] = 0.1

1.724 HA = 0.1

HA = 0.1/ 1.724 = 0.058 M

put this value in 2

[A-] = 0.1-0.058 = 0.042 M

now comming to part C

no of moles of H+ you added = 0.33 M x 0.0059 L = 0.001947 mol

change in pH =

pH = 4.74 + log[0.042 -0.001947 / 0.058 + 0.001947 ]

pH = 4.74 + log[0.040053 / 0.059947]

pH = 4.74 - 0.175

pH = 4.56

change in pH= 4.6 - 4.56

= 0.04

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