assignment 2 solutions

# assignment 2 solutions - Solutions to Assignment 2 Problem...

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Solutions to Assignment 2 Note: This problem is similar to Problem 1-20 on Page 28 of your Segel book. Problem 25 The ionization of H 3 PO 4 is shown below: H + + H 2 PO 4 - H + + HPO 4 = H + + PO 4 H 3 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.: [H + ] [H 2 PO 4 - ] As you see from the equation, for each H there is a H2PO4. Thus you can let each = x. K a = [H 3 PO 4 ] 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.) K a = 7.59 x 10 -3 M = x 2 0.2M - x 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 = 0 Use the quadratic equation to solve for x. x = 34.94 x 10 -3 M = [H + ] pH = - log [H + ] = - log .03494 M = 1.457

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Problem 25 (continued) Part b) KH 2 PO 4 is the salt of the first intermediate ion of phosphoric acid. Therefore the pH = pK a1 + pK a2 2 Look up the pK a s in Appendix IV 2.12 + 7.21 pH = 2 Part c) K 2 HPO 4 is the salt of the second intermediate ion of phosphoric acid. Therefore the = 4.67 pK a2 + pK a3 7.21 + 12.32 pH = 2 = 2 Part d) In this part of the problem you will use the Kb equation as follows to determine the hydrogen ion concentration and then calculate the pH from that.: K b = = 9.76 [HPO 4 = ][OH - ] PO 4 For each OH there is a HPO 4 so you can let x equal each. The last pK
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assignment 2 solutions - Solutions to Assignment 2 Problem...

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