{"id":13283,"date":"2015-12-03T18:20:21","date_gmt":"2015-12-03T18:20:21","guid":{"rendered":"http:\/\/www.oracletutoring.ca\/blog\/?p=13283"},"modified":"2015-12-03T18:20:21","modified_gmt":"2015-12-03T18:20:21","slug":"chemistry-ph","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/chemistry-ph\/","title":{"rendered":"Chemistry:  pH"},"content":{"rendered":"<h1>The tutor introduces and defines pH, an important concept in chemistry.<\/h1>\n<p>Most people know that neutral pH is 7.  However, what is pH really, and what is it based on?<\/p>\n<p>pH is defined as follows:<\/p>\n<p style=\"text-align:center\">pH = -log[H<sup>+<\/sup>]<\/p>\n<p>Here&#8217;s an example:<\/p>\n<p><strong>Find the pH of a 0.0035M solution of HNO<sub>3<\/sub><\/strong><\/p>\n<p>Solution:<\/p>\n<p>First, we write the dissociation equation:<\/p>\n<p style=\"text-align:center\">HNO<sub>3<\/sub>\u2192H<sup>+<\/sup>(aq)+NO<sub>3<\/sub><sup>&#8211;<\/sup>(aq)<\/p>\n<p>Since HNO<sub>3<\/sub> is a strong acid, it completely dissociates.  Thus, [H<sup>+<\/sup>]=0.0035M.<\/p>\n<p style=\"text-align:center\">pH = -log[H<sup>+<\/sup>] = -log(0.0035) = 2.45593 or 2.5 (sig. digits).<\/p>\n<p>HTH:)<\/p>\n<p>Source:<\/p>\n<p>Hebden, James A.  <u>Chemistry:  Theory and Problems, Book Two<\/u>.  Toronto:  McGraw-Hill<br \/>&nbsp;&nbsp; Ryerson Limited, 1980.<\/p>\n<p>Jack of <a href=\"https:\/\/www.oracletutoring.ca\">Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The tutor introduces and defines pH, an important concept in chemistry. Most people know that neutral &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/www.oracletutoring.ca\/blog\/chemistry-ph\/\"> <span class=\"screen-reader-text\">Chemistry:  pH<\/span> Read more \u2192<\/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":[1234,1231,1228,1233,1232,1230,1229],"class_list":["post-13283","post","type-post","status-publish","format-standard","hentry","category-chemistry","tag-logh","tag-hno3","tag-ph","tag-ph-definition","tag-ph-of-nitric-acid-solution","tag-strong-acid","tag-h"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/13283","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=13283"}],"version-history":[{"count":16,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/13283\/revisions"}],"predecessor-version":[{"id":13299,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/13283\/revisions\/13299"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=13283"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=13283"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=13283"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}