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pOH=-log[OH-] Strong Bases- Strong bases are Group 1 or group 2 hydroxides. Strong bases is pretty much the same as strong acids EXCEPT you'll be calculating a pOH first, then going to the pH.
pH + pOH = 14 pH = 14- pOH
Calculate the pH of a 0.0010 M solution of NaOH. NaOH==> Na+ + OH- Therefore, the [OH-] equals 0.0010 M. So, to solve it, you write: pOH = - log (0.0010) = 3.00 pH = 14.000 - 3.00 = 11.00 Calculate the pH of a 0.025 M solution of Ca(OH)2 . Ca(OH)2 ==> Ca2+ + 2OH- Therefore, the [OH-] equals 0.050 M. For every Ca(OH)2 you produce 2OH-. pOH = - log (0.050) = 1.70 pH = 14.000 - 1.70 = 12.30 |
