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Question Number 9346 by An2812 last updated on 02/Dec/16

pH = 14 − (1/2) ( pK_b   −  lgC_b  )

$$\mathrm{pH}\:=\:\mathrm{14}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\left(\:\mathrm{pK}_{\mathrm{b}} \:\:−\:\:\mathrm{lgC}_{\mathrm{b}} \:\right) \\ $$

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