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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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Limblesreap3r Limblesreap3r
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11 years ago
A 0.00474 mol sample of HY is dissolved in enough H2O to form 0.888 L of solution. If the pH of the solution is 2.64, what is the Ka (dissociation constant) of HY?
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#1
11 years ago
initial concentration HY = 0.00474 mol/ 0.888 L= 0.00534 M

[H+]= 10^- 2.64 = 0.00229 M = [Y-]
[HY]= 0.00534 - 0.00229 = 0.00305

Ka = [H+][Y-] / [HY]= (0.00229)^2 / 0.00305 =1.72 x 10^-3
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