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Sigrid O'Kon

Feb 20, 2025

Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.100 M in HClO4(aq) at 25 °C.?

Calculate [H3O ], [ClO4–], and [OH–] in an aqueous solution that is 0.100 M in HClO4(aq) at 25 °C.[H3O ]=?M[ClO4–]=?M[OH–] =?M
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HClO4 is a strong acid that completely ionizes into H3O+ and ClO4-.HClO4 + H2O ==> H3O+ + ClO4-So [H3O+] = [ClO4-] = 0.100 M[OH-][H3O+] = 1 x 10^-14 [OH-] = (1 x 10^-14 / [H3O+]) = (1 x 10^-14 / 0.100) = 1 x 10^-13 M... Show More
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Al is right: "HClO4 -----------------> H+ + ClO4- (H+ = H3O+) on condition that HClO4 dissociates a hundred%, the concentration of H+ and ClO- will the two be 0.a hundred and fifty five mol/L so which you have the concentration of [H3O+] = [ClO4-] = 0.a hundred and fifty five mol/L" you will desire to understand that the concentration of OH^-a million could be calculated from Kw. Kw = [H3O+][OH-] At 25 deg C the value of Kw = 1x10^-14 consequently [HO-] = 1x10^-14 / 0.a hundred and fifty five = 6.45x10^-14 Taking the destructive log of 1x10^-14 = [H3O+][OH-] yields 14 = pH + pOH... Show More
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