Calculate \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in solutions with the following \(\mathrm{pH}\). (a) \(9.0\) (b) \(3.20\) (c) \(-1.05\) (d) \(7.46\)

The formula for pH is given by \(pH = -\log{[\mathrm{H}^{+}]}\). To find the concentration of \([\mathrm{H}^{+}]\), we need to rearrange this formula to \([\mathrm{H}^{+}] = 10^{-pH}\). Calculate the concentration of \([\mathrm{H}^{+}]\) for each given pH value. (a) For pH = 9.0, \([\mathrm{H}^{+}] = 10^{-9.0} = 1 \times 10^{-9}\mathrm{M}\) (b) For pH = 3.20, \([\mathrm{H}^{+}] = 10^{-3.20} = 6.31 \times 10^{-4}\mathrm{M}\) (c) For pH = -1.05, \([\mathrm{H}^{+}] = 10^{-(-1.05)} = 1.12 \times 10^{1}\mathrm{M}\) (d) For pH = 7.46, \([\mathrm{H}^{+}] = 10^{-7.46} = 3.46 \times 10^{-8}\mathrm{M}\)

The relationship between pH and pOH is given by \(pH + pOH = 14\). We can calculate the pOH using this relationship for each given pH value. (a) For pH = 9.0, \(pOH = 14 - 9.0 = 5.0\) (b) For pH = 3.20, \(pOH = 14 - 3.20 = 10.8\) (c) For pH = -1.05, \(pOH = 14 - (-1.05) = 15.05\) (d) For pH = 7.46, \(pOH = 14 - 7.46 = 6.54\)

The formula for pOH is given by \(pOH = -\log{[\mathrm{OH}^{-}]}\). To find the concentration of \([\mathrm{OH}^{-}]\), we need to rearrange this formula to \([\mathrm{OH}^{-}] = 10^{-pOH}\). Calculate the concentration of \([\mathrm{OH}^{-}]\) for each given pOH value. (a) For pOH = 5.0, \([\mathrm{OH}^{-}] = 10^{-5.0} = 1 \times 10^{-5}\mathrm{M}\) (b) For pOH = 10.8, \([\mathrm{OH}^{-}] = 10^{-10.8} = 1.58 \times 10^{-11}\mathrm{M}\) (c) For pOH = 15.05, \([\mathrm{OH}^{-}] = 10^{-15.05} = 8.91 \times 10^{-16}\mathrm{M}\) (d) For pOH = 6.54, \([\mathrm{OH}^{-}] = 10^{-6.54} = 2.89 \times 10^{-7}\mathrm{M}\) In conclusion, for each given pH value, the concentrations of \([\mathrm{H}^{+}]\) and \([\mathrm{OH}^{-}]\) ions are: (a) pH = 9.0: \([\mathrm{H}^{+}] = 1 \times 10^{-9}\mathrm{M}\) and \([\mathrm{OH}^{-}] = 1 \times 10^{-5}\mathrm{M}\) (b) pH = 3.20: \([\mathrm{H}^{+}] = 6.31 \times 10^{-4}\mathrm{M}\) and \([\mathrm{OH}^{-}] = 1.58 \times 10^{-11}\mathrm{M}\) (c) pH = -1.05: \([\mathrm{H}^{+}] = 1.12 \times 10^{1}\mathrm{M}\) and \([\mathrm{OH}^{-}] = 8.91 \times 10^{-16}\mathrm{M}\) (d) pH = 7.46: \([\mathrm{H}^{+}] = 3.46\times 10^{-8}\mathrm{M}\) and \([\mathrm{OH}^{-}] = 2.89 \times 10^{-7}\mathrm{M}\)