Part A Learning Goal: To understand how buffers use reserves of conjugate acid a
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Question
Part A Learning Goal: To understand how buffers use reserves of conjugate acid and conjugate base to counteract the effects of acid or base addition on pH A beaker with 1.60x102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 5.20 mL of a 0.440 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740 Express your answer numerically to two decimal places. Use a minus ( )sign if the pH has decreased A buffer is a mixture of a conjugate acid-base pair In other words, it is a solution that contains a weak acid and its conjugate base, or a weak base and its conjugate acid. For example, an acetic acid buffer consists of acetic acid, CH3 COOH, and its conjugate base, the acetate ion CH3 COO Because ions cannot simply be added to a solution, the conjugate base is added in a salt form (e.g sodium acetate NaCH3COO). View Available Hint(s) pH- 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: Submit uest Answer Provide Feedback Next> H+CH3 COOCH3 COOH Similarly, any added OH ions will be neutralized by a reaction with the conjugate acid OH CHs COOHCHs CO0H20 This buffer system is described by the Henderson- Hasselbalch equation pH pKa log conjugate base [conjugate acidExplanation / Answer
pH of acidic buffer = pka + log(ch3cooK/ch3cooh)
pH = 5
pka of ch3cooh = 4.74
total no of mol of buffer = 1.6*10^2*0.1 = 16 mmol
no of mol of CH3COO- (conjugate base) = x mmol
no of mol of CH3COOH remains = 16-x mmol
5 = 4.74 + log(x/(16-x))
x = 10.32
no of mol of CH3COO- (conjugate base) = x = 10.32 mmol
no of mol of CH3COOH remains = 16-10.32 = 5.68 mmol
after adddition of NaOH
no of mol of NaOH added = 5.2*0.44 = 2.288 mmol
pH of acidic buffer = pka + log((ch3cooK-NaOH)/(ch3cooh+NaOH))
= 4.74+log((10.32-2.288)/(5.68+2.288))
= 4.743
change in pH(DpH) = 5-4.743 = 0.257