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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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You wish to prepare an HC2H3O2 buffer with a pH of 4.24. If the pKa of is 4.74, what ratio of C2H3O2 /HC2H3O2 must you use?
LenaWriter [7]

To solve this problem, we can use the Henderson-Hasselbalch Equation which relates the pH to the measure of acidity pKa. The equation is given as:<span>

<span>pH = pKa + log ([base]/[acid])                ---> 1</span></span>

Where,

[base] = concentration of C2H3O2 in molarity or moles

<span>[acid] = concentration of  HC2H3O2 in molarity or moles</span>

 

For the sake of easy calculation, let us assume that:

[base] = 1

[acid] = x

<span>
Therefore using equation 1,
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<span>log (1 / x) = - 0.5

1 / x = 0.6065 </span></span>

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3 years ago
A solution of NaF is added dropwise to a solution that is 0.0144 M in Ba 2 . When the concentration of F - exceeds __________ M,
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Answer:

When [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate

Explanation:

Ksp of BaF₂ is:

BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)

Ksp = 1.7x10⁻⁶ = [Ba²⁺] [F⁻]²

The solution will produce BaF₂(s) -precipitate- just when [Ba²⁺] [F⁻]² > 1.7x10⁻⁶.

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1.7x10⁻⁶ = [0.0144M] [F⁻]²

1.18x10⁻⁴ = [F⁻]²

0.0109M = [F⁻]

That means, when [F⁻] exceeds 0.0109M concentration, BaF₂ will precipitate

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