Adding a Strong Acid to a Buffer 3 of 6 Constants | Periodic Table Part A Learni

Question

Adding a Strong Acid to a Buffer 3 of 6 Constants | Periodic Table 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 2.00x102 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 A student adds 6.00 mL of a 0.340 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740 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 NaCH3 COO). Express your answer numerically to two decimal places. Use a minus )sign if the pH has decreased View Available Hint(s) 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 ApH Submit H+CH3 COO-CH3 COOH Provide Feedback Next Similarly, any added OH ions will be neutralized by a reaction with the conjugate acid This buffer system is described by the Henderson- Hasselbalch equation coniugate base conjugate acid]

Explanation / Answer

pH of acidic buffer = pka + log(ch3coo^-/CH3COOH)

pka of CH3COOH = 4.74

no of mol of total buffer = M*V

                           = 0.1*200

               = 20 mmol
no of mol of CH3COO- = x

No of mol of CH3COOH = 20-x

5 = 4.74 + log(x/(20-x))

x = 12.9

no of mol of CH3COO- = x = 12.9 mmol

No of mol of CH3COOH = 20-12.9 = 7.1 mmol

after addition ofHCl

No of mol of HCl added = 6*0.34 = 2.04 mmol

pH of acidic buffer = pka + log((ch3coo^- - HCl)/(CH3COOH+HCl))

                 = 4.74 + log((12.9-2.04)/(7.1+2.04))

                 = 4.815

change in pH = 5 - 4.815 = 0.185

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