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

User Joanq
by
3.3k points

2 Answers

14 votes
14 votes

Let's see

  • 25°C is normal room temperature denotion

pOH


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


\\ \rm\Rrightarrow -log[1.2* 10^(-2)]

  • log(ab)=loga+logb


\\ \rm\Rrightarrow -log1.2-log10^(-2)


\\ \rm\Rrightarrow -0.0791812460476+2


\\ \rm\Rrightarrow 2.0791812460476

  • Approximately 2
User Piotr Nowicki
by
3.4k points
10 votes
10 votes

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
User Meidan Alon
by
2.8k points