(a) From the graph we see: Volume of titrant used to reach first equivalence poi

Question

(a)

From the graph we see:

Volume of titrant used to reach first equivalence point = 10 mL

Volume of titrant used to reach second equivalence point = 20 mL

(b)

At half way of the first equivalence point, we have the following relation:

pH = pKa1

Half way of the first equivalence point in this case occurs at 5 mL. At this point pH is approximately 2.1.

Putting values:

pKa1 = 2.1

So,

Ka1 = 10-2.1

At half way of the second equivalence point, we have the following relation:

pH = pKa2

Half way of the second equivalence point in this case occurs at 15 mL. At this point pH is approximately 7.2.

Putting values:

pKa2 = 7.2

So,

Ka2 = 10-7.2

Hope this helps !

Explanation / Answer

perfomed a titration and the resulting data is shown below. The titrant periment is 0.3M NaoH Answer the following questions regarding the being used in this ex PH 7 a) What is the volume of NaOHl used to reach the finst and second equivalence points? First Determine both K,i and K,: based on the curve above. made to determine the Kaj and Ka valwes b)

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