Question

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.

Answer 1

1)

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

we know the equation

K_w = .[OH^-] X [H₃O⁺]

where, K_w = 1.0 X 10^-14

then

[H₃O⁺] = K_w / [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

K_w = .[OH^-] X [H₃O⁺]

where, K_w = 1.0 X 10^-14

then

[H₃O⁺] = K_w / [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

K_w = .[OH^-] X [H₃O⁺]

where, K_w = 1.0 X 10^-14

then

[H₃O⁺] = K_w / [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