Question #0a80d

1 Answer
May 1, 2017

Answer:

  • 46: #K_(sp) = 1.1 * 10^-38#
  • 47: #x = 1.3 * 10^-4 M#
  • 48: #x = 0.0012 M#

Explanation:

  • #46:

The first step to solving molar solubility problems is always to write down the balanced chemical equation of what's going on.

This would be:
#Ba_3(PO_4)_2 (s) rightleftharpoons 3Ba^(2+) (aq) + 2PO_4^(3-) (aq)#

Therefore the expression for #K_(sp)# would be:

#K_(sp) = [Ba^(2+)]^3[PO_4^(3-)]^2#

Setting up an ICE table would give you:

#K_(sp) = (3x)^3(2x)^2 = 108x^5#

Note: Watch your parentheses when simplifying these expressions

Now, the jump to take with all such problems is to realize that #x# is your molar solubility. In the ICE table, it represents how much of your solid dissolved, and therefore is the molar solubility.

You're given a molar solubility, so all you need to do is plug and chug:

#K_(sp) = 108(1.4 * 10^-8)^5 = 1.1 * 10^-38#

  • #47:

Again, same process of setting up the expression:

#Ag_2CrO_4 (s) rightleftharpoons 2Ag^(+) (aq) + CrO_4^(2-) (aq)#

#K_(sp) = [Ag^(+)]^2[CrO_4^(2-)]#

By ICE Table:
#K_(sp) = (2x)^2(x) = 4x^3#

The #K_(sp)# value for this solid is #9.0 * 10^-12.# Hence:

#4x^3 = 9.0 * 10^-12#
#x = 1.3 * 10^-4 M#

  • #48:

This problem is exactly the same process as the previous one, except with a different reaction:

#PbI_2 (s) rightleftharpoons 2I^(-) (aq) + Pb^(2+) (aq)#

#K_(sp) = [Pb^(2+)][I^(-)]^2#

By ICE Table:
#K_(sp) = (x)(2x)^2 = 4x^3#

The #K_(sp)# value for this solid is #7.1×10^(–9).# Hence:

#4x^3 = 7.1×10^(–9)#
#x = 0.0012 M#

To better understand what is exactly going on here, I'd recommend you watch these two videos.

Video 1 (Introduction to #K_(sp)#)
Video 2 (Examples)

Hope that helps :)