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pH-14-1-2-pK-b-lgC-b-




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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