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Answer :
We start by recalling the definition of [tex]\( pOH \)[/tex]:
[tex]$$
pOH = -\log_{10} \left[ OH^- \right]
$$[/tex]
Given that the hydroxide ion concentration is
[tex]$$
\left[ OH^- \right] = 2.0 \times 10^{-2},
$$[/tex]
we substitute this value into the formula:
[tex]$$
pOH = -\log_{10} \left( 2.0 \times 10^{-2} \right).
$$[/tex]
Recall the logarithm property:
[tex]$$
\log_{10} (a \times 10^b) = \log_{10}(a) + b.
$$[/tex]
Thus, we calculate:
[tex]$$
\log_{10} \left( 2.0 \times 10^{-2} \right) = \log_{10}(2.0) + \log_{10}\left(10^{-2}\right).
$$[/tex]
Since
[tex]$$
\log_{10}\left(10^{-2}\right) = -2,
$$[/tex]
and
[tex]$$
\log_{10}(2.0) \approx 0.301,
$$[/tex]
we have:
[tex]$$
\log_{10} \left( 2.0 \times 10^{-2} \right) \approx 0.301 - 2 = -1.699.
$$[/tex]
Now, applying the negative sign to find the [tex]\( pOH \)[/tex]:
[tex]$$
pOH = -(-1.699) \approx 1.699.
$$[/tex]
Rounded to two decimal places, the [tex]\( pOH \)[/tex] is about [tex]\( 1.70 \)[/tex].
Thus, the correct answer is:
Option B.
[tex]$$
pOH = -\log_{10} \left[ OH^- \right]
$$[/tex]
Given that the hydroxide ion concentration is
[tex]$$
\left[ OH^- \right] = 2.0 \times 10^{-2},
$$[/tex]
we substitute this value into the formula:
[tex]$$
pOH = -\log_{10} \left( 2.0 \times 10^{-2} \right).
$$[/tex]
Recall the logarithm property:
[tex]$$
\log_{10} (a \times 10^b) = \log_{10}(a) + b.
$$[/tex]
Thus, we calculate:
[tex]$$
\log_{10} \left( 2.0 \times 10^{-2} \right) = \log_{10}(2.0) + \log_{10}\left(10^{-2}\right).
$$[/tex]
Since
[tex]$$
\log_{10}\left(10^{-2}\right) = -2,
$$[/tex]
and
[tex]$$
\log_{10}(2.0) \approx 0.301,
$$[/tex]
we have:
[tex]$$
\log_{10} \left( 2.0 \times 10^{-2} \right) \approx 0.301 - 2 = -1.699.
$$[/tex]
Now, applying the negative sign to find the [tex]\( pOH \)[/tex]:
[tex]$$
pOH = -(-1.699) \approx 1.699.
$$[/tex]
Rounded to two decimal places, the [tex]\( pOH \)[/tex] is about [tex]\( 1.70 \)[/tex].
Thus, the correct answer is:
Option B.
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