Unformatted text preview: + ], and [OH] values for a 0.0525M solution of hydroioidic acid. HI is strong and ionizes about 100% so at equilibrium [H+] = 0.0525M so pH = 1.280 then pOH = 12.720 and [OH] = 1.90 x 1013 M 6. Calculate the pH for a 1.35 M solution of a weak acid HA with K a = 8.39 x 105 . This is a weak acid so equilibrium is reached. HA(aq) + H 2 O(l) ' A (aq) + H 3 O + (aq) I 1.35  0 0 C x  +x +x E (1.35x)  (x) (x) K a = 8.39 x 105 = x 2 / (1.35x) approx = x 2 / 1.35 8.39 x 105 (1.35) = x 2 take the square root of both sides x = 0.01064 which is also H+ so pH = log x = 1.973 = pH check approximation 0.01064 / 1.35 x 100% = 0.788% which is less than 5% so valid...
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 Nitric acid, (aq)

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