Buffers Q2 Part A Learning Goal: How many moles of acid are in this buffer solut

Question

Buffers Q2 Part A Learning Goal: How many moles of acid are in this buffer solution? The following calculations are similar to those you will complete as part of the lab. 6.96x 10 moles 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: My Answers Give Up Correct H + CH , COO CH, COOH Similarly, any added OH ions will be neutralized by a reaction with the conjugate acid: Part B How many moles of the conjugate base are in this buffer solution? OH + CH COOH CH, COO + HO 5. 04x10 -3 moles This buffer system is described by the Henderson Hasselbalch equation My Answers Give Up pH = pK+log. [conjugate base [conjugate acid Correct The pK of acetic acid is 4.74. Your Buffer System: Part C A beaker with 1.20x102 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 If you add 5.90 mL of a 0.330 MHC1 solution to the beakerhow much will the ph change? Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased. DVD Aze n o o - ? 46 Submit Hints My Answers Give Up Review Part Incorrect; Try Again Provide feedback Continue

Explanation / Answer

Given that [ CH3COO-] = 5.04 x 10-3 moles

[ CH3COOH] = 6.96 x 10-3 moles

[HCl] = 0.33 M x 0.0059 L= 1.95 x  10-3

pH = pKa + log [CH3COO-]/[CH3COOH] --------(1)

On addition of HCl,

CH3COO- + HCl ------------> CH3COOH + Cl-

Hence,

pH = pKa + log [ CH3COO-] -[HCl] / [CH3COOH] + [HCl] ---------(2)

Since HCl is a strong acid, pH will be decreased compared to original pH.

To get change in pH, subtract (2) from (1).

Then,

Change in pH = pKa + log [CH3COO-]/[CH3COOH] -  {pKa + log [ CH3COO-] -[HCl] / [CH3COOH] + [HCl]}

=  log [CH3COO-]/[CH3COOH] - log [ CH3COO-] -[HCl] / [CH3COOH] + [HCl]

= log [5.04 x10-3] / [6.96x 10-3] - log [5.04x10-3- 1.95 x10-3]/[6.96x10-3-1.95x10-3]

= -0.07

Therefore,

change in pH = -0.07

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