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kykrilka [37]
3 years ago
6

Which of the following is an oxidation-reduction reaction?

Chemistry
1 answer:
KiRa [710]3 years ago
3 0

C₆H₁₂O₆ + 6 O₂ ⇒ 6 CO₂ + 6 H₂O

This is an oxidation-reduction reaction, because oxidation numbers of carbon and oxygen atoms are different than before reaction.

(-_-(-_-)-_-)

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The Henderson-Hasselbalch equation relates the pH of a buffer solution to the pKa of its conjugate acid and the ratio of the con
Kobotan [32]

Answer:

The correct answer is 190.5 mL of 1.00 M KH₂PO₄

Explanation:

A phosphate buffer is composed by phosphate acid (KH₂PO₄) and its conjugated base (K₂HPO₄). To obtain the relation between the concentrations of base and acid to add, we use Henderson-Hasselbach equation:

pH= pKa + log \frac{base}{acid}

We have: pH= 6.97 and pKa= 7.21. So, we replace the values in the equation:

6.97= 7.21 + log \frac{base}{acid}

6.97-7.21= log \frac{base}{acid}

-0.24= log \frac{base}{acid}

10^{-0.24}= \frac{base}{acid}

0.575 = \frac{base}{acid}

\frac{0.575}{1}= \frac{base}{acid}

It means that you have to mix a volume 0.575 times of conjugated base and 1 volume of acid. If we assume a total buffer concentration of 1 M, we have:

base + acid = 1

base= 1 - acid

We replace in the previous equation:

0.575= \frac{1-acid}{acid}

0.575 acid= 1 - acid

0.575 acid + 1 acid= 1

1.575 acid = 1

acid= 1/1,575

acid= 0.635

base= 1 - acid = 1 - 0.635 = 0.365

For a total volume of 300 ml, the volumes of both acid and base will be:

300 ml x 0.635 M = 190.5 ml of acid (KH₂PO₄)

300 ml x 0.365 M= 109.5 ml of base (K₂HPO₄)

We can corroborate our calculations as follows:

190.5 ml + 109.5 ml = 300 ml

109.5 ml / 190.5 ml = 0.575

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