{"id":9838,"date":"2015-04-17T22:54:33","date_gmt":"2015-04-17T22:54:33","guid":{"rendered":"http:\/\/www.oracletutoring.ca\/blog\/?p=9838"},"modified":"2015-04-17T22:54:33","modified_gmt":"2015-04-17T22:54:33","slug":"chemistry-common-ion-effect","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/chemistry-common-ion-effect\/","title":{"rendered":"Chemistry: common ion effect"},"content":{"rendered":"

The tutor tackles an essential concept of grade 12 chemistry.<\/h1>\n

I introduced solubility product back in my Dec 1<\/a> post. Today the question is slightly different: How much of a salt will dissolve in a solution that already contains one of its ions?<\/p>\n

Example 1<\/strong>: What is the solubility of lead(II)chloride in a 0.150 M solution of NaCl?<\/p>\n

Solution:<\/p>\n

The dissociations of NaCl and PbCl2<\/sub> are as follows:
\n
\nNaCl(s) \u2192 Naaq<\/sub>+<\/sup> + Claq<\/sub>–<\/sup>
\nPbCl(s) \u2192 Pbaq<\/sub>2+<\/sup> + 2Claq<\/sub>–<\/sup>
\n<\/span><\/p>\n

The solubility product of PbCl2<\/sub> is given by
\n
\nksp<\/sub>=[Pb2+<\/sup>][Cl–<\/sup>]2<\/sup>=1.2×10-5<\/sup><\/span><\/p>\n

Let the concentration of PbCl2<\/sub> in moles per litre be given by x. Then the following ion concentrations are present:<\/p>\n\n\n\n
ion<\/td>\nPb2+<\/sup><\/td>\nCl–<\/sup><\/td>\n<\/tr>\n
concentration<\/td>\nx<\/td>\n2x+0.150  (0.150 from NaCl)<\/td>\n<\/tr>\n<\/table>\n

Putting the concentrations into the ksp<\/sub> expression gives
\n
\nksp<\/sub>=x(2x+0.150)2<\/sup>=1.2×10-5<\/sup><\/p>\n

Now, we can make the approximation that 2x+0.150\u22480.150. Hebden (see source below) argues as follows: First, realize that 1.2×10-5<\/sup>=0.000012. If x(2x+0.150)2<\/sup>=0.000012, then x<0.000012\/(0.150)2<\/sup>. Therefore, x<0.000533; it follows that 2x< 0.00107.\n\nIf 2x < 0.00107, then 2x + 0.150 < 0.00107 + 0.150 = 0.151 in significant digits. The 2x term is worth, at most, 0.001\/0.150 or 0.7%; the approximation that 2x+0.150\u22480.150 is valid.\n\nApplying the approximation, we arrive at\n\nx(2x+0.150)2<\/sup>\u2248x(0.150)2<\/sup>=1.2×10-5<\/sup><\/span><\/p>\n

We continue<\/p>\n

x(0.150)2<\/sup>=1.2×10-5<\/sup><\/span><\/p>\n

x=1.2×10-5<\/sup>\/(0.150)2<\/sup>
\nx=0.00053
\n<\/span>
\nApparently, the solubility of PbCl2<\/sub> in 0.150M NaCl is 0.00053 or 5.3×10-4<\/sup>M. Perhaps next post I’ll compute its solubility in pure water by way of comparison.<\/p>\n

Source:<\/p>\n

Hebden, James A. Chemistry: Theory and Problems, Book Two<\/em>. Toronto: McGraw-Hill   Ryerson Limited, 1980.<\/p>\n

Jack of Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.<\/p>\n","protected":false},"excerpt":{"rendered":"

The tutor tackles an essential concept of grade 12 chemistry. I introduced solubility product back in my Dec 1 post. Today the question is slightly different: How much of a salt will dissolve in a solution that already contains one …<\/p>\n

Chemistry: common ion effect<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[11],"tags":[706,707,708,317],"class_list":["post-9838","post","type-post","status-publish","format-standard","hentry","category-chemistry","tag-common-ion-effect","tag-common-ion-effect-approximation","tag-solubility","tag-solubility-product"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/9838","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/comments?post=9838"}],"version-history":[{"count":36,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/9838\/revisions"}],"predecessor-version":[{"id":9874,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/9838\/revisions\/9874"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=9838"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=9838"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=9838"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}