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NeX [460]
3 years ago
13

Acetic acid (HC2H3O2, Ka=1.8x10^-5) is a weak acid. Calculate the pH of an aqueous solution of .25M acetic acid.

Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
5 0

Answer: The pH of an aqueous solution of .25M acetic acid is 2.7

Explanation:

HC_2H_3O_2\rightarrow H^+C_2H_3O_2^-

 cM              0             0

c-c\alpha        c\alpha          c\alpha  

So dissociation constant will be:

K_a=\frac{(c\alpha)^{2}}{c-c\alpha}

Give c= 0.25 M and \alpha = ?

K_a=1.8\times 10^{-5}

Putting in the values we get:

1.8\times 10^{-5}=\frac{(0.25\times \alpha)^2}{(0.25-0.25\times \alpha)}

(\alpha)=0.0084

[H^+]=c\times \alpha

[H^+]=0.25\times 0.0084=0.0021

Also pH=-log[H^+]

pH=-log[0.0021]=2.7

Thus pH is 2.7

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What happens to the energy that is lost when water freezes?
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What is the molarity of a solution that contains 14 moles of solute and 2 liters of solution?
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  7 M

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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:

9.34x10^-4

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

Mass of PbCl2 = 0.2393 g

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

Molarity of PbCl2 =?

Molarity = mole /Volume

Molarity of PbCl2 = 8.61x10^-4/0.05

Molarity of PbCl2 = 0.01722 M

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]

[Pb^2+] = 0.0159 M

[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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