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The problem gives that the solubility of Silver dichromate is $8.3\times10^{-3} g/100mL$. I need to find $K_{sp}$ which is supposed to be $2.8\times10^{-11}$.

$K_{sp}$ is $\ce{[Ag+]^2[Cr2O7^{2-}]}$

I changed the $8.3\times 10^{-3}$ from g/mL to mol/L and got $1.9\times 10^{-8}$.

Silver dichromate breaks into 2 Ag and one dichromate.

So that would be $(2 \times 1.9\times10^{-8})^2 \times ( 1.9\times10^{-8})$

That's what I did but it's not giving me the right answer. Where am I going wrong here?

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The problem gives that the solubility of Silver dichromate is $\mathrm{8.3\cdot10^{-3} g/100\,mL}$.

[…]

I changed the $8.3\cdot10^{-3}$ from g/mL to mol/L and got $1.9\cdot10^{-8}$.

[…] Were am I going wrong here?

Was the solubility given in $\mathrm{g/100\,mL}$?

$$\mathrm{8.3\cdot10^{-3} g/100\,mL = 8.3\cdot10^{-2}\,g/L = \frac{8.3}{431.72}\,mol/L = 1.92\cdot10^{-4}\,mol/L}$$

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    $\begingroup$ Thanks. I was doing something stupid going from mL to L. $\endgroup$ – Caesium-133 Mar 26 '15 at 6:58
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    $\begingroup$ My pleasure! Blame it on the late night or the early morning ;) $\endgroup$ – Klaus-Dieter Warzecha Mar 26 '15 at 6:59
  • $\begingroup$ Haha, yea. Think I pull way too many of those (late nights). I'm lucky the internet makes it easy t find a fresh set of eyes to give a second look. $\endgroup$ – Caesium-133 Mar 26 '15 at 7:08
  • $\begingroup$ Yes, I'm not on CST. It's 08:00 over here. $\endgroup$ – Klaus-Dieter Warzecha Mar 26 '15 at 7:11
  • $\begingroup$ It's just about 3:30am here. Kind of the only time it's quiet in here so... $\endgroup$ – Caesium-133 Mar 26 '15 at 7:20

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