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Rainbow [258]
4 years ago
6

A solution is prepared by mixing the 50 mL of 1 M NaH2PO4 with 50 mL of 1 M Na2HPO4. On the basis of the information, which of t

he following species is present in the solution at the lowest concentration?Extra ContentH3PO4 <--> H+ + H2PO4- Ka1 = 7.2 E-3H2PO4- <--> H+ + HPO42- Ka2 = 6.3 E-8HPO42- <--> H+ + PO43- Ka3 = 4.5 E-13answer options:A. PO43-B. H2PO4-C. HPO42-D. Na+
Chemistry
1 answer:
barxatty [35]4 years ago
4 0

Answer:

A. PO43-

Explanation:

in the given is buffer . so H2PO4- and HPO42- both are present with equal concentration . Na+ is spectator ion it is also present in the concentration higher than the given species above .

but PO4-3 is not present . so it is lowest concentration

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Answer:

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The true absorbance for a 1.0 x 10 −5 M solution is 0.7526. If the percentage stray light for a spectrophotometer is 0.56%, calc
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Answer:

The percentage deviation is  \Delta M = 1.87%

Explanation:

From the question we are told that  

     The concentration is of the solution is C = 1.0*10^{-5} M

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      The percentage of transmittance due to stray light z = 0.56% =\frac{0.56}{100}  = 0.0056

Generally Absorbance is mathematically represented as

           A = -log T

Where T is  the percentage of true transmittance

    Substituting value  

          0.7526 = - log T

              T = 10^{-0.7526}

                  = 0.177

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The Apparent absorbance is mathematically represented

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           A_p = -log(0.177 + 0.0056)

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The percentage by which apparent absorbance deviates from known absorbance is mathematically evaluated as

       \Delta A = \frac{A -A_p}{A} * \frac{100}{1}

              = \frac{0.7526 - 0.7385}{0.7526} * \frac{100}{1}

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Since Absorbance varies directly with concentration the percentage deviation of the apparent concentration from know concentration  is

              \Delta M = 1.87%

           

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