Exercise 15.26 Determine the pH of each solution. Part A 2.5×10 2 M HI Express y

Question

Exercise 15.26

Determine the pH of each solution.

Part A

2.5×102 M HI

Express your answer to two decimal places.

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Part B

0.113 M HClO4

Express your answer to three decimal places.

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Part C

a solution that is 5.3×102 M in HClO4 and 3.0×102 M in HCl

Express your answer to two decimal places.

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Part D

a solution that is 1.90% HCl by mass (Assume a density of 1.01 g/mL for the solution.)

Express your answer to three decimal places.

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Exercise 15.26

Determine the pH of each solution.

Part A

2.5×102 M HI

Express your answer to two decimal places.

pH =

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Part B

0.113 M HClO4

Express your answer to three decimal places.

pH =

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Part C

a solution that is 5.3×102 M in HClO4 and 3.0×102 M in HCl

Express your answer to two decimal places.

pH =

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Part D

a solution that is 1.90% HCl by mass (Assume a density of 1.01 g/mL for the solution.)

Express your answer to three decimal places.

pH =

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Explanation / Answer

A)

HI --> H+ + I-

we know that

HI is a strong acid

so 100 % dissociation

so

[H+] = [HI]

[H+] = 2.5 x 10-2

now

we know that

pH = -log [H+]

so

pH = -log 2.5 x 10-2

pH = 1.602

B)

HCl04 is the strongest acid known

So 100 % dissociation

[H+] = [HCl04] = 0.113

now

pH = -log [H+]

pH = -log 0.113

pH = 0.947

C)

now

HCl04 and HCl are both strong acids

both contribute H+ to the solution

[H+] from HCl04 = 5.3 x 10-2

[H+] from HCl = 3 x 10-2

so

total [H+] = 8.3 x 10-2

now

pH = -log [H+]

pH = -log 8.3 x 10-2

pH = 1.081

D)

consider 100 ml of solution

then

mass of solution = density x volume

mass of solution = 1.01 x 100

mass of solution = 101 g

now

% mass = mass of HCl x 100 / mass of solution

1.9= mass of HCl x 100 / 101

mass of HCl = 1.919 g

now

moles = mass / molar mass

also

molar mass of HCl = 36.46 g

so

moles of HCl = 1.919/ 36.46

moles of HCl = 0.052633

now

concentration = moles x 1000 / volume (ml)

concentration of HCl = 0.052633 x 1000 / 100

concentration of HCl = 0.52633

now

as HCl is a strong acid

[H+]= [HCl] = 0.5263

now

pH = -log [H+]

pH = -log 0.5263

pH = 0.279

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