Solution Assn. 2

# Solution Assn. 2 - Solutions to Assignment 2 Problem 25 The...

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Unformatted text preview: Solutions to Assignment 2 Problem 25 The ioonization of H 3 PO 4 is shown below: H 3 PO 4 H + + H 2 PO 4- H + + HPO 4 = H + + PO 4 Part a) In this part of the problem you will use the K a equation as follows to determine the hydrogen ion concentration and then calculate the pH from that.: K a = [H + ] [H 2 PO 4- ] [H 3 PO 4 ] As you see from the equation, for each H there is a H2PO4. Thus you can let each = x. K a = 7.59 x 10-3 M = x 2 0.2M - x Look up the first pK a for phosphoric acid in Appendix IV of the Segel book. (It is 2.12) and then take the antilog of -2.12 to get the Ka, which is 7.59 x 10-3 M. (Since the concentration of the acid is given as .2 M and it is less than 100 x the K a , you will have to include the removal of x in the denominator of the equation.) Note: This problem is similar to Problem 1-20 on Page 28 of your Segel book. x 2 = 1.52 x 10-2- 7.59 x 10-3 x x 2 + 7.59 x 10-3 x – 1.59 x !0-2 = Use the quadratic equation to solve for x. x = 34.94 x 10-3 M = [H + ] pH = - log [H + ] = - log .03494 M = 1.457 Problem 25 (continued) Part b) KH 2 PO 4 is the salt of the first intermediate ion of phosphoric acid. is the salt of the first intermediate ion of phosphoric acid....
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## This note was uploaded on 02/21/2009 for the course BIS BIS102 taught by Professor Hilt during the Winter '08 term at UC Davis.

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Solution Assn. 2 - Solutions to Assignment 2 Problem 25 The...

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