Another CHemistry Problem..plz

Pre-College Issues High School Life
1

<li><p>At 10 degrees Celcius, 8.9 x 10^-5 grams of AgCl dissolves in 100 mL of water.</p>

<p>i.) Write teh equation for the dissasociation of AgCl.
ii.) Calculate the solubility of AgCl in water
iii.) Calculate the value of Ksp for AgCl</p></li>
</ol>

<p>b.) At 25 degrees Celcius, the value of Ksp for PbCl2 is 1.6x10^-5 and Ksp for AgCl is 1.8x10^-10</p>

<pre><code> i) If 60mL of .04 M NaCl is added to 60 mL of .03 M Pb(NO3)2, will a precipitate form?

ii) Calculate the equilibrium value of Pb2 in 1 Liter of saturated PbCl2 solution to .25 mole NaCl.

iii.) If .1 M NaCl is added to beaker containing both .12 M AgNO3 and .15 M Pb(NO3)2 at 25 degree celcius, which will precipate first?
</code></pre>

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<p>Ahhh, this thread reminds me why I dropped Adv. Chem.</p>

3

<p>i) AgCl => Ag^+ + Cl^1-</p>

<p>ii) Find the number of moles of AgCl in 8.9x10^-5 g of AgCl. Then you divide by .1 L to find molarity. Since AgCL is ionic, it will dissociate completely, and whatever your molarity is your solubility constant. </p>

<p>iii) Use Ksp = [Ag+][Cl1-], plugging in the solubility constant value and solve for ksp.</p>

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<p>b) i) Calculate the number of moles of Cl1-, and divide that by .120 L since that's the total amount of solution. Calculate the moles of Pb2+ and do the same. Then you use the equation, Q= [Cl1-]^2[Pb2+], and plug in those values you found for each. Cl1- is being squared since PbCl2 has two parts of Cl1-. Once you solve for Q, compare to Ksp. </p>

<p>ughhh...too much chem. anyways, it was good review</p>

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<p>So in problem 1 #iii, Ag and Cl are equal?</p>

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<p>yea they are jason</p>

<p>btw jason, post in like 1 spot dude. we get you need the answers, but they come fast enough on the AP tests board.</p>

<p>i mean you posted there and then PMed me the same question.......
then you posted this there, then posted it here (or vice-versa, whatever)</p>