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

Answer:

pH = 1.046

[OH⁻] = 1.11 ×10⁻¹³ M

pOH = 12.954

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, $1.12\times 10^{-13}M$, 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.

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

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

$[OH^-]=1.12\times 10^{-13}M$

The $OH^-$ concentration is, $1.12\times 10^{-13}M$