Question

Determine the [OH-], pH, and pOH of a solution with a [H+] of 0.090 M at 25 °C. [OH-] = pH = pOH = 0 POH =

Answer 1

pH = 1.046

[OH⁻] = 1.11 ×10⁻¹³ M

pOH = 12.954

Step-by-step explanation:

Given: [H⁺] = 0.090 M = 9 ×10⁻² M; T = 25°C

As, pH = - log [H⁺]

⇒ pH = - log (9 ×10⁻²) = 1.046

The self-ionisation constant of water is given by ,

Kw = [H⁺] [OH⁻]

and, pKw = pH + pOH

Since at room temperature, 25°C: Kw = 1.0 × 10⁻¹⁴ , pKw = 14

∴ Kw = [H⁺] [OH⁻] = 1.0 × 10⁻¹⁴

⇒ [OH⁻] = (1.0 × 10⁻¹⁴) ÷ [H⁺] = (1.0 ×10⁻¹⁴) ÷ [9 ×10⁻²] = 0.111 ×10⁻¹² = 1.11 ×10⁻¹³ M

And,

pH + pOH = pKw = 14

⇒ pOH = 14 - pH = 14 - 1.046 = 12.954

Answer 2

Answer : The concentration of
`OH^-` ion, pH and pOH of solution is,
`1.12* 10^(-13)M`, 1.05 and 12.95 respectively.

Explanation : Given,

Concentration of
`H^+` ion = 0.090 M

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

The expression used for pH is:

`pH=-\log [H^+]`

First we have to calculate the pH.

`pH=-\log [H^+]`

`pH=-\log (0.090)`

`pH=1.05`

The pH of the solution is, 1.05

Now we have to calculate the pOH.

`pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-1.05=12.95`

The pOH of the solution is, 12.95

Now we have to calculate the
`OH^-` concentration.

`pOH=-\log [OH^-]`

`12.95=-\log [OH^-]`

`[OH^-]=1.12* 10^(-13)M`

The
`OH^-` concentration is,
`1.12* 10^(-13)M`