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GalinKa [24]
3 years ago
6

Calculate the ph of a 0.20 m solution of iodic acid (hio3, ka = 0.17).

Chemistry
1 answer:
saveliy_v [14]3 years ago
7 0
Iodic acid partially dissociates into H+ and IO3- 
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will  be released as that of H+, its concentration is also X.  The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3]; 
<span>Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; </span>
<span>At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; </span>
<span>so 0.17 = x² / (0.20 - x); </span>
<span>Solving for x using the quadratic formula: </span>
<span>x = [H+] = 0.063 M or pH = - log [H+] = 1.2.</span>
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