91Ó°ÊÓ

Use the formula \(\mathrm{pH}+\mathrm{pOH} = 14\). Subtract the pOH value from 14 to determine the pH value. a. \(pH = 14 - pOH = 14 - 4.95 = 9.05\) b. \(pH = 14 - pOH = 14 - 7.00 = 7.00\) c. \(pH = 14 - pOH = 14 - 12.94 = 1.06\) d. \(pH = 14 - pOH = 14 - 1.02 = 12.98\)

Use the formula \(\mathrm{pH}=-\log_{10}([\mathrm{H^+}])\), rearranged to \([\mathrm{H^+}] = 10^{-\mathrm{pH}}\). Plug in the pH value for each case and calculate the hydrogen ion concentration. a. \([\mathrm{H^+}] = 10^{-9.05} = 8.91 \times 10^{-10} \,\mathrm{mol/L}\) b. \([\mathrm{H^+}] = 10^{-7.00} = 1.00 \times 10^{-7} \,\mathrm{mol/L}\) c. \([\mathrm{H^+}] = 10^{-1.06} = 8.69 \times 10^{-2} \,\mathrm{mol/L}\) d. \([\mathrm{H^+}] = 10^{-12.98} = 1.04 \times 10^{-13} \,\mathrm{mol/L}\) The hydrogen ion concentration for each pOH value is as follows: a. \(8.91 \times 10^{-10} \,\mathrm{mol/L}\) b. \(1.00 \times 10^{-7} \,\mathrm{mol/L}\) c. \(8.69 \times 10^{-2} \,\mathrm{mol/L}\) d. \(1.04 \times 10^{-13} \,\mathrm{mol/L}\)