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-log(2.3 X 10^-3 M)

= 2.6

14 - 2.6 = 11.4 pH ( OH - )

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13y ago
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4mo ago

The pH level when [OH-] equals 2.3 x 10^-3 M can be calculated using the equation: pOH = -log[OH-]. Given [OH-] = 2.3 x 10^-3 M, pOH = -log(2.3 x 10^-3) ≈ 2.64. Since pH + pOH = 14, pH = 14 - pOH = 14 - 2.64 ≈ 11.36.

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Q: PH level when OH- equals 2.3 x 10-3 M?
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