A typical vitamin C tablet (containing pure ascorbic acid, H2C6H6O6) weighs 500. mg. One vitamin C tablet is dissolved in enough water to make 200.0 mL of solution. calculate the pH of this solution. Ascorbic acid is a diprotic acid.

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Given that 500. mg of Ascorbic acid (&amp;nbsp;C6H6O6H2) is dissolved in water to make 200.0mLsolution.Molarity of ascorbic acid==Mass&amp;nbsp;of&amp;nbsp;ascorbic&amp;nbsp;acid&amp;nbsp;(g)Molar&amp;nbsp;mass&amp;nbsp;of&amp;nbsp;ascorbic&amp;nbsp;acid&amp;nbsp;(g/mol)×1000=0.5g176.13&amp;nbsp;g/mol×1000200mL=0.014MC6H6O6H2(aq)+H2O(!)□&amp;nbsp;&amp;nbsp;&amp;nbsp;C6H6)O6H-(aq)+H3O*(aq)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Ka1=7.9×10-5Initial: 0.014M 0 0Change: -x x xEquilibrium: 0.014-x x xKa1=[C6H6O6H-][H3O+][C6H6O6H2]7.9×10-5=x×x0.014-x(0.014-x) can be taken as 0.014x=1.051×10-3C6H6O6H-(aq)+H2O(l)□&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;C6H6O62-(aq)+H3O+(aq)&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;Ka2=1.6×10-12Initial:&amp;nbsp;1.051×10-3m0&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;1.051×10-3MChange: -y y yEquilibrium:&amp;nbsp;(1.051×10-3-y)Ka2=(1.051×10-3+y)×y1.051×10-3-y1.6×10-12=1.051×10-3×y1.051×10-3y=1.6×10-12M[H+]&amp;nbsp;of the solution= x+y=(1.051×10-3&amp;nbsp;M)&amp;nbsp;+&amp;nbsp;(1.6×10-12&amp;nbsp;M)=1.05116×10-3&amp;nbsp;MpH&amp;nbsp;=&amp;nbsp;-log[H+]=&amp;nbsp;-log&amp;nbsp;(1.05116×10-3&amp;nbsp;M)=&amp;nbsp;2.97∪&amp;nbsp;3.0

A typical vitamin C tablet (containing pure ascorbic

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