Question

An aqueous solution has a hydroxide ion concentration of 1.20 × 10–2 M at 25°C. Calculate the pOH of the solution.

Answer 1

Let's see

• 25°C is normal room temperature denotion

pOH

$\\ \rm\Rrightarrow -log[OH^-]$

$\\ \rm\Rrightarrow -log[1.2\times 10^{-2}]$

• log(ab)=loga+logb

$\\ \rm\Rrightarrow -log1.2-log10^{-2}$

$\\ \rm\Rrightarrow -0.0791812460476+2$

$\\ \rm\Rrightarrow 2.0791812460476$

• Approximately 2

Answer 2

Given↷

• [OH-] = 1.20 × 10^–2 M
• Temperature = 25°

To find↷

• pOH of the solution

Answer ↷

• pOH = 1.92

Solution ↷

• [OH-] = 1.20 × 10^–2 M
• pOH = - log [OH^1-]
• pOH = - log [1.20 × 10^–2]
• pOH = 1.92