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Calculate the [OH−] for 0.0055 M H2SO4. Answer in units of M. Calculate the [H3O+] for 0.011 M Ca(OH)2. Answer in units of M. Please explain the steps.

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[H3O+] and [OH-]....... Sulfuric acid is a strong acid, but only for the first H+ which comes off. The remaining HSO4^- ion is a weak acid and its dissociation is negligible in the presence of 0.0055M H+. Therefore, the [H+] of a 0.0055M solution of H2SO4 is 0.0055M. Kw = [H+][OH-] = 1.00x10^-14 .............. at about 25C [OH-] = Kw / [H+] = 1.00x10^-14 / 0.0055 = 1.8x10^-12 You may not have been exposed to acid/base equilibria, and know little about strong and weak acids and chemical equilibria. Therefore, your teacher may be expecting you to assume that H2SO4 produces two H+ for each molecule. In which case, double the H+ concentration and cut the OH- concentration in half. Calcium hydroxide is a strong base, but not a very soluble base. Therefore, what little Ca(OH)2 dissolves will completely ionize. A quick check shows that a saturated solution of Ca(OH)2 at 20C has a concentration of 0.0233M. Therefore, you 0.011M solution is viable. Ca(OH)2(aq) --> Ca2+ + 2OH- [OH-] = 0.022M Kw = [H+][OH-] = 1.00x10^-14 [H+] = 1.00x10^-14 / 0.022 = 4.5x10^-13M

            
        

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