Chem 162-2011 chapter 15 2nd half homework problems

Chem 162-2011 chapter 15 2nd half homework problems - Chem...

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Chem 162-2010 Hill & Petrucci Homework Chapter 15 (Note that my algebra-solving calculator requires my setting up equations involving “,X” and lots of parentheses. Ignore these equations. Also, in reviewing the homework, it seems that some of the problems might have been solved more easily if the Henderson-Hasselbalch equation had been used.) More Acid-Base Equilibria 81. A solution is 0.405 M HCOOH (formic acid) and 0.326 M in the salt HCOONa (sodium formate). What is the pH of this buffer solution? HCOOH + H 2 O H 3 O + + HCOO - HCOOH + H 2 O H 3 O + + HCOO - Initial 0.405 M 0.326M Change Equilibrium HCOOH + H 2 O H 3 O + + HCOO - Initial 0.405 M 0 0.326M Change -X +X +X Equilibrium 0.405 - X +X 0.326 + X K a = 1.8 x 10 -4 Small K, so drop the X’s. Henderson-Hasselbalch: pH = pK a + log([A - ]/[HA]) pH = -log(1.8 x 10 -4 ) + log([0.326]/[0.405]) = 3.65 83. What mass of (NH 4 ) 2 SO 4 (s) must be dissolved in 0.100 L of 0.350 M NH 3 (aq) to produce a buffer solution with pH = 10.05? NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O NH 4 + + OH - Initial 0.350M Y Change Equilibrium pH = 10.05 Henderson-Hasselbalch: pH = pK a + log([A - ]/[HA]) K b = K NH3 = 1.8 x 10 -5 K ca = (1 x 10 -14 )/(1.8 x 10 -5 ) = 5.56 x 10 -10 pH = -log(5.56 x 10 -10 ) + log([0.350]/[Y]) 10.05 = -log(5.56 x 10 -10 ) + log([0.350]/[Y]) (((10.05) = ((-log(5.56 x 10 -10 )) + (log([0.350]/[Y])))),X) Y = 0.056M NH 4 +
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MW (NH 4 ) 2 SO 4 = 132.1gmol -1 (NH 4 ) 2 SO 4 2NH 4 + + SO 4 2- 0.056molNH 4 + /L NH 4 + x (1 NH 4 ) 2 SO 4 /2 NH 4 + ) x 0.100L x 132.1 g/mol = 0.37g (NH 4 ) 2 SO 4 (s) 85. If 1.00 mL of 0.250 M HCl is added to 50.0 mL of the buffer solution described in Problem 81, what will be the pH of the final solution? From problem 81: pH = 3.65; therefore [H 3 O + ] = 2.24 x 10 -4 HCOOH = 0.405M; HCOO - = 0.326M HCOOH + H 2 O H 3 O + + HCOO - Initial Change Equilibrium 0.405 2.24 x 10 -4 0.326 0.250 mol/L x 0.001L = 0.000250 mol HCl added to 0.0500 L. Therefore 0.0050 mol HCl added to 1L. Therefore, new concentration of H 3 O + is 2.24 x 10 -4 + 0.0050 = 5.224 x 10 -3 M H 3 O + . Since more strong acid was added to the system at equilibrium, the reaction will shift to the left while going to completion. Reaction goes to completion because reactions of strong acids go to completion. HCOOH + H 2 O H 3 O + + HCOO - Initial 0.405 (5.224 x 10 -3 ) 0.326 Change +(5.224 x 10 -3 ) -(5.224 x 10 -3 ) -(5.224 x 10 -3 ) Equilibrium 0.410 0 0.321 Henderson-Hasselbalch: pH = pK a + log([A - ]/[HA]) pH = (-log(1.8 x 10 -4 )) + (log([0.321]/[0.410])) pH = 3.64 Acid-Base Indicators 89. Why are so many more acid-base indicators suitable for the titration of a strong acid with a strong base than are suitable for the titration of a weak acid with a strong base? In comparing figures 15.15 and 15.16 one can see that the inflection part of the curve for the strong acid and strong base covers a pH range from approximately 3.5 to 11.5, whereas the inflection part of the curve for the weak acid and strong base covers a pH range of only approximately 7.0 to 11.0. There are more indicators available for 3.5 to 11.5 then for the limited 7.0 to 11.0. 93. A 0.100 M HCl(aq) solution contains thymol blue indicator and has a red color. A 0.100 M NaOH(aq) solution with phenolphthalein indicator also has a red color. What should be the color when equal volumes of the two solutions (with their indicators) are mixed?
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Chem 162-2011 chapter 15 2nd half homework problems - Chem...

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