In an aqueous solution at $25^\circ C$ has twice as many $OH^-$ as pure water its $\ce{pOH}$ will be$:$
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We know that pure water at $25^\circ C$ has $10^{−7}M$ of $OH^−.$
According to the problem, the aqueous solution at the same temperature has $2 \times 10^{−7}\ce{M OH^−}$
Therefore, its $\ce{pOH} = −\log(2\times 10^{−7})$
$\ce{pOH} = 6.6989$
or, $\ce{pOH} = 6.699$
According to the problem, the aqueous solution at the same temperature has $2 \times 10^{−7}\ce{M OH^−}$
Therefore, its $\ce{pOH} = −\log(2\times 10^{−7})$
$\ce{pOH} = 6.6989$
or, $\ce{pOH} = 6.699$
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