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den301095 [7]
3 years ago
14

Which solution has the lowest pH? A. 1.0 M HNO2 B. 1.0 M CH3COOH C. 1.0 M HCOOH D. 1.0 M HNO3 E. 1.0 M HPO4–

Chemistry
2 answers:
Svetlanka [38]3 years ago
8 0
The stronger the acid the lower the pH, because pH = log { 1 / [H+]}. You can discard easily the options B and C because they are organic acid which are relatively weak. To compare properly the inorganic acids you should use a Ka or pKa table. You can predict that HPO4 (2-) is less strong because the third dissociation is weaker the first. So, you should only need to search for the Ka or pKa of HNO2 and HNO3. Every one who is familiar with the stronger acids knows that HNO3 is one of the strongest acid, stronger indeed than HNO2. So, the strongest acid of the list is HNO3 and it has the lowest pH. <span>Answer: option D. 1.0 M HNO3.</span>
lianna [129]3 years ago
7 0

Answer:

Answer: option D. 1.0 M HNO3.

Explanation:

#platolivesmatter

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onsider the reversible dissolution of lead(II) chloride. P b C l 2 ( s ) − ⇀ ↽ − P b 2 + ( a q ) + 2 C l − ( a q ) PbClX2(s)↽−−⇀
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Answer:

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Explanation:

Step 1:

The balanced equation for the reaction.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

Step 2:

Data obtained from the question:

Mass of PbCl2 = 0.2393 g

Volume = 50mL

concentration of Pb^2+, [Pb^2+] = 0.0159 M

Concentration of Cl^-, [Cl^-] = 0.0318 M

Equilibrium constant, Kc =?

Step 3:

Determination of the number of mole PbCl2.

The number of mole of PbCl2 can be obtained as follow:

Molar Mass of PbCl2 = 207 + (35.5x2) = 278g/mol

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Number of mole =Mass /Molar Mass

Number of mole of PbCl2 = 0.2393/278 = 8.61x10^-4 mole

Step 4:

Determination of Molarity of PbCl2.

At this stage we shall obtain the molarity of PbCl2. This is shown below:

Mole of PbCl2 = 8.61x10^-4 mole

Volume = 50mL = 50/1000 = 0.05L

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Step 5:

Determination of the equilibrium constant Kc.

PbCl2( s ) <=> Pb^2+(aq) + 2Cl^−(aq)

The equilibrium constant Kc for the equation above is given by:

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

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[Cl^-] = 0.0318 M

[PbCl2] = 0.01722 M

Kc =?

Kc = [Pb^2+] [Cl^-]^2 / [PbCl2]

Kc = 0.0159 x (0.0318)^2/ 0.01722

Kc = 9.34x10^-4

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