Calculate the pH at which Mg(OH)2 begin to precipitate from solution containing 0.1M Mg2+ ions. Ksp of Mg(OH)2= 1*10^-11.
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I'm getting 8.64. (s is solubility)
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Wait.I think I did wrong.. I didn't even use the 0.1 molar...
Ignore this answer.
OK, answer is 9. I didn't even read the question properly.. Basically, I'll tell you the steps here but my answer above is wrong anyways. Now, ksp= [Mg] [OH] ^2=0.1[OH]^2. If you solve, you'll get OH concentration as 10^-5. So, H+ concentration is 10^-14/10^-5=10^-9. pH is negative log of that which gives 9.
Ahhh this one is confusing
I will ask the same question again
OK good.
Please answer correctly
I will.
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