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dusya [7]
3 years ago
9

The student plans to conduct a spectrophotometric analysis to determine the concentration of Cu2+(aq) in a solution. The solutio

n has a small amount of Co(NO3)2 (aq) present as a contaminant. The student is given a diagram below, which shows the absorbance curves for aqueous solutions of Co2+ (aq) and Cu2+ (aq)
c. The spectrophotometer available to the student has a wavelength range of 400 nm to 700 nm. What wavelength should the student use to minimize the interference from the presence of the Co2+ (aq)?

Chemistry
1 answer:
Lelechka [254]3 years ago
6 0

Answer:

The wavelength the student should use is 700 nm.

Explanation:

Attached below you can find the diagram I found for this question elsewhere.

Because the idea is to minimize the interference of the Co⁺²(aq) species, we should <u>choose a wavelength in which its absorbance is minimum</u>.

At 400 nm Co⁺²(aq) shows no absorbance, however neither does Cu⁺²(aq). While at 700 nm Co⁺²(aq) shows no absorbance and Cu⁺²(aq) does.

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3 years ago
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?
olga_2 [115]

Answer:

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,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[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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