Answer is not -.46, -.07 or .04. Part C Learning Goal: The following calculation

Question

Answer is not -.46, -.07 or .04.

Part C

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 pH = pK_a + log [conjugate base]/[conjugate acid] Your Buffer System: A break with 1.20 times 10^2 of an acetic acid buffer with a pH.60 is sitting on a bench top The total molarity of acid and conjugate base in this buffers is 0.100 M. How many moles of acid are in this buffer solution? How many moles of the conjugate base are in this buffer solution? If you add 5.90 mL of a 0.330 M HCl 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.

Explanation / Answer

pH = pKa + log [salt / acid]

      = 4.74 + log [5.04 x 10^-3 /6.96 x 10^-3]

      = 4.60

pH = 4.60

moles of acid = 6.96 x 10^-3

moles of conjugate base = 5.04 x 10^-3

moles of HCl (C) = 5.90 x 0.330 / 1000 = 1.947 x 10^-3

pH = pKa + log [conjugate base - C / acid + C ]

     = 4.74 + log [5.04 x 10^-3 - 1.947 x 10^-3 / 6.96 x 10^-3 + 1.947 x 10^-3]

     = 4.28

pH = 4.28

pH change = 4.28 - 4.60 = -0.32

pH change = - 0.32

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