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lina2011 [118]
2 years ago
8

Help please acids and basis

Chemistry
1 answer:
Marizza181 [45]2 years ago
6 0

Let's see

  • [OH-] be x

\\ \rm\Rrightarrow pOH=-log[OH-]

\\ \rm\Rrightarrow pOH=-logx

\\ \rm\Rrightarrow 4.86=-logx

\\ \rm\Rrightarrow logx=-4.86

\\ \rm\Rrightarrow x=10^{-4.86}

\\ \rm\Rrightarrow x=1.38M

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