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Brrunno [24]
3 years ago
6

Acetic acid has a pKa of 4.74. Buffer A: 0.10 M HC2H3O2, 0.10 M NaC2H3O2 Buffer B: 0.30 M HC2H3O2, 0.30 M NaC2H3O2 Buffer C: 0.5

0 M HC2H3O2, 0.10 M NaC2H3O2 Which of the above buffers has the highest buffer capacity? [ Select ] Which of the above buffers has the lowest buffer capacity? [ Select ]
Chemistry
1 answer:
e-lub [12.9K]3 years ago
6 0

Answer:

Buffer B has the highest buffer capacity.

Buffer C has the lowest buffer capacity.

Explanation:

An effective weak acid-conjugate base buffer should have pH equal to pK_{a} of the weak acid. For buffers with the same pH, higher the concentrations of the components in a buffer, higher will the buffer capacity.

Acetic acid is a weak acid and CH_{3}COO^{-} is the conjugate base So, all the given buffers are weak acid-conjugate base buffers. The pH of these buffers are expressed as (Henderson-Hasselbalch):

pH=pK_{a}(CH_{3}COOH)+log\frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}

pK_{a}(CH_{3}COOH)=4.74

Buffer A:    pH=4.74+log(\frac{0.10}{0.10})=4.74

Buffer B:   pH=4.74+log(\frac{0.30}{0.30})=4.74

Buffer C:   pH=4.74+log(\frac{0.10}{0.50})=4.04

So, both buffer A and buffer B has same pH value which is also equal to pK_{a} . Buffer B has higher concentrations of the components as compared to buffer A, Hence, buffer B has the highest buffer capacity.

The pH of buffer C is far away from pK_{a} . Therefore, buffer C has the lowest buffer capacity.

     

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Consider the titration of 10.00 mL of a monoprotic weak acid with 0.1234 M NaOH. If the equivalence point volume of NaOH was det
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The initial concentration of the unknown acid is 0.1900 M.

Explanation:

Titration is a chemical method of analysis to know the concentration and volume of the unknown chemical or analyte.

The formula for the titration is:

Macid x Vacid = Mbase x V base

The volume must be in litres. The volume is given in ml it should be divided with 1000 to obtain values in litre.

Data given are:

volume of acid= 10 ml 0.01 L

Molarity of the acid = ?

volume of the NaOH or base = 15.4 ml or 0.0154 L (equivalence point of the base)

molarity of the base = 0.1234 M

Applying the formula and putting the values, we get

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