Question

What is the pH of a solution of KOH with a hydroxide concentration of [OH⁻] = 1.10 x 10⁻⁴

Answer 1

Answer:

$p[H+] = 10.042$

Explanation:

As we know that

$pKw = pH + pOH$......eq (1)

we will calculate the pH of OH- and then we will calculate the pH of H+

So p[OH-] $= - log [1.10 * 10^{-4}]$

Solving the right side of the equation, we get

p[OH-]

$= - [-3.958]\\= 3.958$

Now we know that

$pKw = 14.0$

Substituting the value of pOH in the above equation, we get -

$14.0 = p[H+] + 3.958\\p[H+] = 14 - 3.958\\p[H+] = 10.042$