A solution has a [H^+] of 1.0times 10^-5M 6. What is its [OH^-] 7. What is its pH? 8. What is its POH? A solution has a [OH^-] of 3.6times 10^-7M 9. What is its [H^+] 10. What is its pH? 11. What is its pOH? A solution has a [H^+] of 5.6times 10^-6M 12. What is its [OH^-] ?

6. The concentration of hydroxide ions, [OH-], can be calculated using the ion product of water, Kw, which is 1.0 x 10^-14 at 25°C. The formula is [OH-] = Kw / [H+]. Substituting the given [H+] value, we get [OH-] = 1.0 x 10^-14 / 1.0 x 10^-5 = 1.0 x 10^-9 M.<br /><br />7. The pH of a solution is calculated using the formula pH = -log[H+]. Substituting the given [H+] value, we get pH = -log(1.0 x 10^-5) = 5.<br /><br />8. The pOH of a solution is calculated using the formula pOH = -log[OH-]. Substituting the calculated [OH-] value, we get pOH = -log(1.0 x 10^-9) = 9.<br /><br />9. The concentration of hydrogen ions, [H+], can be calculated using the ion product of water, Kw, which is 1.0 x 10^-14 at 25°C. The formula is [H+] = Kw / [OH-]. Substituting the given [OH-] value, we get [H+] = 1.0 x 10^-14 / 3.6 x 10^-7 = 2.78 x 10^-8 M.<br /><br />10. The pH of a solution is calculated using the formula pH = -log[H+]. Substituting the calculated [H+] value, we get pH = -log(2.78 x 10^-8) = 7.56.<br /><br />11. The pOH of a solution is calculated using the formula pOH = -log[OH-]. Substituting the given [OH-] value, we get pOH = -log(3.6 x 10^-7) = 6.44.<br /><br />12. The concentration of hydroxide ions, [OH-], can be calculated using the ion product of water, Kw, which is 1.0 x 10^-14 at 25°C. The formula is [OH-] = Kw / [H+]. Substituting the given [H+] value, we get [OH-] = 1.0 x 10^-14 / 5.6 x 10^-6 = 1.79 x 10^-8 M.