Find the pH and percent ionization of each HF solution. (Ka for HF is 6.8×10−4.)a.  Find the pH of a 0.290 M HF solution.Express your answer to two decimal places.b. Find the percent dissociation of a 0.290 M HF solution.Express your answer using two significant figures. c. Find the pH of a 0.110 M HF solution.Express your answer to two decimal places.d. Find the percent dissociation of a 0.110 M HF solution.Express your answer using two significant figures. e. Find the pH of a 5.00×10−2 M HF solution. Express your answer to two decimal places.f. Find the percent dissociation of a 5.00×10−2 M HF solution. Express your answer using two significant figures.

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Asked Oct 23, 2019
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Find the pH and percent ionization of each HF solution. (Ka for HF is 6.8×10−4.)

a.  Find the pH of a 0.290 M HF solution.

Express your answer to two decimal places.

b. Find the percent dissociation of a 0.290 M HF solution.

Express your answer using two significant figures.
 
c. Find the pH of a 0.110 M HF solution.

Express your answer to two decimal places.

d. Find the percent dissociation of a 0.110 M HF solution.
Express your answer using two significant figures.
 
e. Find the pH of a 5.00×10−2 M HF solution. 
Express your answer to two decimal places.
f. Find the percent dissociation of a 5.00×10−2 M HF solution. 
Express your answer using two significant figures.
 
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Expert Answer

Step 1

The pH of a solution is related to the hydrogen ion concentration of a solution. For a solution of weak acid, the hydrogen ion concentration is determined on the basis of its dissociation constant.

The pH of a solution is given by expression (1), in which, [H+] is the hydrogen ion concentration of the solution.

The percent dissociation of a species present in the solution is calculated by dividing the composition of the dissociated species (cd) by the composition of the undissociated species (cu), and then multiplying it with 100.

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pH log H (1) % dissociation ax100 (2) C

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Step 2

a)

 

The amount of HF undergoes dissociation in the given solution is calculated by its dissociation constant expression and ICE table.

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H F HF 00 Initial 0.290 Change +x +x -X Equilibrium 0.290-x х х [н[F K HF (х)(x) (0.290-x) -x2-0.00068x + 0.000197 = 0 a 6.8x 10 x -0.0144, 0.0137

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Step 3

The value of concentration of H+ and F- ions can never be negative. Therefore, the positive value 0.0137 wi...

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рH -- 1og(0.0137) =1.9

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