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Answer :
The hydronium ion concentration [H3O+] in a 0.010 M solution of Sr(OH)2 is calculated using the water ion product (Kw) and the molarity of OH-. The concentration of H3O+ is found to be 5.0 x 10^-14 M.
When Sr(OH)2 dissolves in water, it divides into Sr2+ ions and OH- ions. Because one formula unit of Sr(OH)2 generates two OH- ions, the molarity of OH- in the solution is double the original molarity of Sr(OH)2. In this case, [OH-] is 2(0.010 M) = 0.020 M.
To find the Hydronium ion concentration [H3O+], we can use the product of the concentrations of H3O+ and OH-, known as the water ion product (Kw), which equals 1.0 x 10^-14 at 298 K. By rearranging the equation for Kw ([H3O+] [OH-] = Kw), we can calculate [H3O+] as Kw / [OH-]. Plugging in our values, we get [H3O+] = (1.0 x 10^-14) / 0.020 M = 5.0 x 10^-14 M.
The concentration of H3O+ in a 0.010 M solution of Sr(OH)2 is 5.0 x 10^-14 M.
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