Question

Compute the corresponding (OH^-) or (H₃O⁺) in each of the following conditions. Describe the condition as acidic or base.

1. (H₃O⁺) = 3.7 x 10^-10 M

2. (OH^-) = 2.4 x 10^-13 M

Answer 1

we know that

$\star$$[OH-][H3O+]=1 \times10^{-14}$

$\star P^H=-log[H^+]$

$\star P^{OH}=-log[OH^-]$

1)

$[H_3O^+]$ =3.7 x 10^-10 M

corresponding [OH-]=2.7 x10^-5

$P^H=-log[H^+]=-log[3.7\times10^{-10}]$

=9.44 (>7)

thus, basic.

2)[OH^-] = 2.4 x 10^-13 M

corresponding [H⁺]=4.1 x 10^-2

$P^H=-log[H^+]=-log[4.1\times10^{-2}]$

=1.387 (<7)

thus ,acidic