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« previous | 6 of 19 next » Adding a Strong Acid to a Buffer Part A Learning Goal To understand how buffer conjugate acid and conjugate base to counteract the effects of acid or base addition on pH A beaker with 105 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 4.20 mL of a 0.360 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740 s use reserves of 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 toa solution, the conjugate base is added in a salt form (e.g., sodium acetate NaCH3 COO) Hints 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 My Answers Give Up Provide Feedback Continue H+ + CH3COO-CH3COOH Similarly, any added OH ions will be neutralized by a reaction with the conjugate acid OH + CH3COOH CH3COO-+ H2O This buffer system is described by the Henderson- Hasselbalch equation conjugate base conjugate acidExplanation / Answer
The pH of acetic acid buffer = 5.0 and pKa = 4.74
We know pH = pKa + log [conjugate base]/[acid] from Hendersen equation
5.00 = 4.74 + log x/0.1-x
x = 0.064
Now
A- + H+ -------------> HA
0.064x105=6.72 0 0.036x105=3.78 initial mmoles
- 4.2 x0.36 =1.512 - change
5.208 0 5.292 after
Thus
pH = 4.74 + log 5.208/5.292
= 4.733
The change in pH = 0.007
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