This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 3]/[H 2 CO 3 ] = [H + ] 2 /[H 2 CO 3 ] = 106.4 and [H 2 CO 3 ] + [H + ] = 0.0001M If [H 2 CO 3 ] >> [H + ] then [H 2 CO 3 ] = 0.0001M [H + ] 2 = 106.4 x 0.0001 = 4.0 x 1011 [H + ] = 6.3 x 106 = 105.2 1 1b pH = log (5.2) = 5.2 2 1c The percent ionized may be calculated by dividing the hydrogen ion concentration by the concentration of the solution and multiplying by 100. % ionized = (105.2 )/(104 ) = 101.2 = 0.06 x 100 = 6% 2 1d At a concentration of 0.01M, pH = 4.2 % ionized = (104.2 )/(102 ) = 102.2 = 0.006 x 100 = 0.6% 2 1e In general we may conclude that there is more ionization in a dilute solution. Numbers to the left in red are the possible points....
View
Full
Document
This document was uploaded on 03/29/2011.
 Spring '09

Click to edit the document details