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Katyanochek1 [597]
3 years ago
6

Why do buffer systems use either a weak acid or a weak base?

Chemistry
2 answers:
uysha [10]3 years ago
7 0

Answer:

Weak acids and weak bases are partially dissociate

Explanation:

I took the test and got 100%

mr_godi [17]3 years ago
3 0

Answer:

Weak acids and weak bases are only partially dissociated.

Explanation:

When a small amount of an acid or a base is added to water (or a solution of some strong acid or base of some certain pH \!,) the pH of that solution tends to change significantly.

In contrast, when the same amount of acid or base is added to a buffer system of some certain pH\!, the pH of this system tends to change by a much smaller amount.

Weak acids and weak bases dissociate only partially. That allows the buffer solutions to contain a large reservoir of both the (undissociated) weak acid (or weak base) and the corresponding conjugate ion, while maintaining the required pH.

This reservoir of the partially-dissociated weak acid (or weak base) and its conjugate ion allows the buffer system to absorb \rm H^{+} and \rm OH^{-} that were added to the solution without much change to the pH.

For example, consider a 1\; \rm L buffer solution that initially included 0.1\; \rm mol of acetic acid \rm CH_3COOH (a weak acid with pK_\text{a} = 4.765) and 0.1\; \rm mol\! of sodium acetate \rm CH_3COONa. The pH of this solution would be 4.765.

In contrast, if a solution of \rm HCl (a strong acid) and an equal amount of \rm NaCl also needs to be 4.765, only 10^{-4.765}\; \rm mol \approx 0.0000172\; \rm mol of each species could be included.

Assume that 0.001\; \rm mol of {\rm NaOH}\, (s) is added to the buffer solution of \rm CH_3COOH and \rm CH_3COONa. The quantity of the partially-dissociated weak acid in this solution is much larger than that of the \rm OH^{-} added to the solution.

Only a small portion of the weak base would be consumed.The pH of the solution would likely change by less than 0.001.

In contrast, assume that 0.001\; \rm mol of {\rm NaOH}\, (s) is added to the 1\; \rm L solution with 0.0000172\; \rm mol of \rm HCl (fully-dissociated) and \rm NaCl.

All the \rm HCl\! in this solution would be consumed. The pH of this solution would go well above 7.

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