{"id":13324,"date":"2015-12-05T16:53:29","date_gmt":"2015-12-05T16:53:29","guid":{"rendered":"http:\/\/www.oracletutoring.ca\/blog\/?p=13324"},"modified":"2017-09-07T17:23:52","modified_gmt":"2017-09-07T17:23:52","slug":"chemistry-h-from-ph","status":"publish","type":"post","link":"https:\/\/www.oracletutoring.ca\/blog\/chemistry-h-from-ph\/","title":{"rendered":"Chemistry:  [H+] from pH"},"content":{"rendered":"<h1>The tutor explores an implication of the definition of pH.<\/h1>\n<p>In my <a href=\"?p=13283\">Dec 3<\/a> post I showed the definition of pH and how to calculate it. What if you know the pH, but want the [H<sup>+<\/sup>]?<\/p>\n<p><strong>Example 1: Find the [H<sup>+<\/sup>] of a solution with pH 5.3<\/strong><\/p>\n<p>Solution:<\/p>\n<p>From the definition of pH<\/p>\n<p>pH=-log[H+]<\/p>\n<p>it follows that, after multiplying both sides by -1,<\/p>\n<p>-pH=log[H+]<\/p>\n<p>Next, we take the antilog (likely <span style=\"font-variant: small-caps;\">shift<\/span> log or 2nd log on the calculator):<\/p>\n<p>antilog(-pH)=[H+]<\/p>\n<p>In this case, with pH=5.3,<\/p>\n<p>antilog(-5.3)=[H+]<\/p>\n<p>and the calculator yields<\/p>\n<p>5.0&#215;10^(-6)=[H+]<\/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 Ryerson, 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 explores an implication of the definition of pH. In my Dec 3 post I showed the definition of pH and how to calculate it. What if you know the pH, but want the [H+]? Example 1: Find the &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"more-link\" href=\"https:\/\/www.oracletutoring.ca\/blog\/chemistry-h-from-ph\/\"> <span class=\"screen-reader-text\">Chemistry:  [H+] from pH<\/span> Read More &raquo;<\/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":[1240,1228,1239],"class_list":["post-13324","post","type-post","status-publish","format-standard","hentry","category-chemistry","tag-antilog","tag-ph","tag-h-from-ph"],"_links":{"self":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/13324","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=13324"}],"version-history":[{"count":9,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/13324\/revisions"}],"predecessor-version":[{"id":23433,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/posts\/13324\/revisions\/23433"}],"wp:attachment":[{"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/media?parent=13324"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/categories?post=13324"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.oracletutoring.ca\/blog\/wp-json\/wp\/v2\/tags?post=13324"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}