In: Chemistry

# 1. Calculate [H3O+] in the following aqueous solution at 25 C: [OH−]=1.1×10−9 M. 2. Calculate [H3O+]...

1. Calculate [H3O+] in the following aqueous solution at 25 C: [OH−]=1.1×10−9 M.

2. Calculate [H3O+] in the following aqueous solution at 25 C: [OH−]=2.4×10−2 M .

3. Calculate [H3O+] in the following aqueous solution at 25 C: [OH−]=6.2×10−12 M

then classify the solutions as acidic or basic.

## Solutions

##### Expert Solution

1)

.[OH-] = 1.1 X 10-9 M

we know the equation

Kw = .[OH-] X [H3O+]

where, Kw = 1.0 X 10-14

then

[H3O+] = Kw / [OH-]

substitute the value

[H3O+] = 1.0 X 10-14/ 1.1 X 10-9 = 9.09 X 10-6 M

[H3O+] = 9.09 X 10-6 M

pH = - log[H3O+] = -log(9.09 X 10-6 ) = 5.04

Solution is acidic because pH from 0 to 7 is acidic

2)

.[OH-] = 2.4 X 10-2 M

we know the equation

Kw = .[OH-] X [H3O+]

where, Kw = 1.0 X 10-14

then

[H3O+] = Kw / [OH-]

substitute the value

[H3O+] = 1.0 X 10-14/ 2.4 X 10-2 = 4.167 X 10-13 M

[H3O+] = 4.167 X 10-13 M

pH = - log[H3O+] = -log(4.167 X 10-13  ) = 12.38

Solution is basic because pH from 7 to 14 is basic

3)

.[OH-] = 6.2 X 10-12 M

we know the equation

Kw = .[OH-] X [H3O+]

where, Kw = 1.0 X 10-14

then

[H3O+] = Kw / [OH-]

substitute the value

[H3O+] = 1.0 X 10-14/ 6.2 X 10-12 = 1.61 X 10-3 M

[H3O+] = 1.61 X 10-3​​​​​​​ M

pH = - log[H3O+] = -log(1.61 X 10-3 ) = 2.79

Solution is acidic because pH from 0 to 7 is acidic

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