Question

Determine the H*. OH , and pOH of a solution with a pH of 9.57 at 25 °C. [OH] pOH

Answer 1

we have

pH = 9.57

We know the formulas

pH = -log[H⁺]

pOH = -log[OH^-]

pH + pOH = 14

Now

pH = 9.57 = -log[H⁺]

log[H⁺] = -pH = -9.57

hence

[H⁺] = 10^-pH = 10^-9.57 =

So

[H⁺] = 2.69 $\times$ 10^-10 M

now

pH + pOH = 14

hence

pOH = 14 - pH = 14 - 9.57 = 4.43

pOH = 4.43

Now

pOH = -log[OH^-]

log[OH^-] = -pOH

hence

[OH^-] = 10^-pOH = 10^-4.43 =

[OH^-] = 3.71 $\times$ 10^-5 M

So answers are

[H⁺] = 2.69 $\times$ 10^-10 M

[OH^-] = 3.71 $\times$ 10^-5 M

pOH = 4.43