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What is the pH of a 0.025 M solution of KOH?(A) 1.60 (B) 3.69 (C) 10.31 (D) 12.40
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`KOH rarr (K^+) +(OH^-)`
`P^OH = -log [OH^-]= -log[0.025] = 1.6`
To find` P^H ` we need to use the equation;
`P^Kw = P^H +P^(OH)`
Assuming that the temperature is 25 C;
`14= P^H +1.6`
`P^H = 14 - 1.6 =12.4`
So the `P^H` of the solution is 12.4.
Posted by jeew-m on August 16, 2013 at 3:32 AM (Answer #1)
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