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

Respuesta :

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

Let's see

  • 25°C is normal room temperature denotion

pOH

[tex]\\ \rm\Rrightarrow -log[OH^-][/tex]

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

  • log(ab)=loga+logb

[tex]\\ \rm\Rrightarrow -log1.2-log10^{-2}[/tex]

[tex]\\ \rm\Rrightarrow -0.0791812460476+2[/tex]

[tex]\\ \rm\Rrightarrow 2.0791812460476[/tex]

  • Approximately 2

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