<p>Can somebody please help me with this problem? My textbook doesn't explain this at all, seriously.</p>
<p>Find the solubility of AgSCN (Ksp = 1.2 x 10^-12) in 0.50 M NH3 if Kd = 4 x 10^-8 for Ag(NH3)2. Keep two significant figures.</p>
<p>Thank you for any help.</p>
<p>what is kd anyway?</p>
<p>and y not Ag(NH4) ???sorry I dont get the question</p>
<p>Kd is apparently the reciprocal of Kf. So, Kd = [Ag +][NH3]^2 / [Ag(NH3)2 +].</p>
<p>The problem says Ag(NH3)2+. I guess when it asks for solubility, it means the concentration of Ag+. I have another Kd problem that I can't figure out, but I thought this one is probably the easier.</p>
<p>I know some of you are really tempted to do this problem. Go ahead, indulge yourself please.</p>
<p>haha, sorry, I took chem last year, so last night I was going through my lastyear notebook (hah, I still keep them around), so it will take a while for me to figure it out. If someone already have the solution, plz go ahead.</p>