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VikaD [51]
3 years ago
6

What characteristic frequencies in the infrared spectrum of your sodium borohydride reduction product will you look for to deter

mine whether the carbonyl group (in ethyl vanillin) has been converted to an alcohol functional group? That is, if the transformation has occurred, which characteristic frequencies should be present in the infrared spectrum, and which should be absent? How would the infrared spectrum for the sodium borohydride reduction product be affected if the product were still wet with water? Explain your reasoning.
Chemistry
1 answer:
kkurt [141]3 years ago
5 0

Answer:

A)The characteristic frequency to look out for is  1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.

B)Characteristic frequency  present in the infrared spectrum will be at a peak of  3300-3400 cm-1 which will be due to O-H stretch.

C)If the product is wet with water there will be no change in the infrared spectrum

Explanation:

The characteristic frequency to look out for is  1720-1740 cm-1 (for C=O) for which will disappear in the end product but initially present in the reactant.

Characteristic frequency  present in the infrared spectrum will be at a peak of 3300-3400 cm-1 which will be due to O-H stretch.

If the product is wet with water there will be no change in the infrared spectrum

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If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
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pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

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[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

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