Asked by DrBob222
Look on your calculator. You know the definition of pH. It is pH=-log(H^+). Well, we have the same thing in pKa, pKb, pOH, pKw, etc etc.
pKsp = -log Ksp.
Punch 57.685 into the calculator. Now change the sign to make it negative. Now look for a button that says 10<sup>x</sup> and punch that. The answer is something like ??? x 10^-58.
Try it. You'll like it.
A really weird salt of the type A3B4 is put into water and the saturated solution forms. For this salt, pKsp = 57.685. What is the base-10 log of the solubility? (Need logarithm for answer)
Can someone give me some steps to follow?
is it 2.065e-58 ?
If so, how do I give that answer as a logarithm? This might sound kind of silly, but I haven't done logs in awhile... but does that mean the answer needs to contain "log" of something?
Thanks
Yes, that's the answer.
If pK<sub>sp</sub>= 57.685, then
K<sub>sp</sub>= 2.065 x 10<sup>-58</sup>.
The "logarithm" is in the formula for the conversion.
pKsp = -log Ksp.
Punch 57.685 into the calculator. Now change the sign to make it negative. Now look for a button that says 10<sup>x</sup> and punch that. The answer is something like ??? x 10^-58.
Try it. You'll like it.
A really weird salt of the type A3B4 is put into water and the saturated solution forms. For this salt, pKsp = 57.685. What is the base-10 log of the solubility? (Need logarithm for answer)
Can someone give me some steps to follow?
is it 2.065e-58 ?
If so, how do I give that answer as a logarithm? This might sound kind of silly, but I haven't done logs in awhile... but does that mean the answer needs to contain "log" of something?
Thanks
Yes, that's the answer.
If pK<sub>sp</sub>= 57.685, then
K<sub>sp</sub>= 2.065 x 10<sup>-58</sup>.
The "logarithm" is in the formula for the conversion.
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