The concentration of hydronium ions and hydroxide ions can vary widely in a solution, easily from 1 M to 1 x 10-14 M. This variation is too wide to fit on an analog readout. In order to make this into a smaller scale suitable for an analog readout, the log to the base 10 was used so that the read out is focused on powers of 10. Note that the log [1] = 0, and the log of 1 x 10-14 = -14. It was then decided that the numbers should be positive so the sign was changed. The way to measure concentrations of both H+ and OH- was thus defined as -log. This was then referred to as p. Thus, pH = -log[H+], pOH = -log[OH-], and pK = -log[Keq]. You need to figure out how to do this mathematical manipulation on your calculator. The acid-base reaction in pure water is H2O + H2O ⇔ H3O+1 + OH-1 and it was found that in pure water under standard conditions the [H3O+1] = [OH-1] = 1 x 10-7 M Neutral pH is defined as having a [H+] or [H3O+1] of 1 x 10-7 M. The pH of a neutral solution is thus pH = -log[1 x 10-7] = The pOH of a neutral solution is thus pOH = -log[1 x 10-7] =
The concentration of hydronium ions and hydroxide ions can vary widely in a solution, easily from 1 M to 1 x 10-14 M. This variation is too wide to fit on an analog readout. In order to make this into a smaller scale suitable for an analog readout, the log to the base 10 was used so that the read out is focused on powers of 10. Note that the log [1] = 0, and the log of 1 x 10-14 = -14. It was then decided that the numbers should be positive so the sign was changed. The way to measure concentrations of both H+ and OH- was thus defined as -log. This was then referred to as p. Thus, pH = -log[H+], pOH = -log[OH-], and pK = -log[Keq]. You need to figure out how to do this mathematical manipulation on your calculator. The acid-base reaction in pure water is H2O + H2O ⇔ H3O+1 + OH-1 and it was found that in pure water under standard conditions the [H3O+1] = [OH-1] = 1 x 10-7 M Neutral pH is defined as having a [H+] or [H3O+1] of 1 x 10-7 M.
The pH of a neutral solution is thus pH = -log[1 x 10-7] =
The pOH of a neutral solution is thus pOH = -log[1 x 10-7] =
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