Answers Practice Problems Chap. 1 PDF

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Practice Problems, Chapter 1

1) What is the concentration of H+ in a solution of 0.1 M NaOH? NaOH is a strong base, which means it all dissociates in solution. Thus, [OH-] = [NaOH] = 0.1M [H+][OH-(0.1M)] = 1 x 10-14 [H+] = 1 x 10-14 / 0.1M = 1 x 10-13 2) What is the concentration of OH- in a solution in which the H+ concentration is 0.00013 M? [H+][OH-] = 1 x 10-14 1 x 10-14 / 0.00013M = 7.7 x 10-11 3) What is the pH of 3 x 10-2 M KOH? KOH is a strong base, thus [OH] = 3 x 10-2 [H+][OH-] = 1 x 10-14 1 x 10-14 / 3 x 10-2M = 3.3 x 10-13 pH = 12.5 4) Calculate the pKa of lactic acid, given that when the concentration of lactic acid is 0.010 M and the concentration of lactate is 0.087 M, the pH is 4.80. pH = pKa + log [A- ] [HA] 4.80 = pKa + 0.94

4.80 = pKa + log 0.087M 0.010M 4.80 – 0.94 = pKa

pKa = 3.86

5) Calculate the pH of a mixture of 0.1 M acetic acid and 0.2 M sodium acetate. The pKa of acetic acid is 4.76. pH = pKa + log [A- ] [HA]

pH = 4.76 + log 0.2M 0.1M

pH = 4.76 + 0.30 = 5.06

6) You have a solution acetic acid at pH 5.30. What is the ratio of the concentrations of acetate and acetic acid? The pKa of acetic acid is 4.76. pH = pKa + log [A- ] [HA] 100.54 = 10 log[A-]/[HA] 3.5 = [A- ] [HA]

5.30 = 4.76 + log [A-] 0.54 = log [A-] [HA] [HA]

7) Determine the value of Go, in Kilojoules (kJ), for the reaction below at 25o C, and indicate whether NH4NO3 spontaneously dissolves in water at room temperature. NH4NO3(s) + H2O(l)

NH4+ (aq) + NO3- (aq)

Ho = 28,050 Joules So = 108.7 Joules/Kelvin Go = Ho - TSo

= 28,050 Joules - (25 + 273 Kelvin)(108.7 Joules/Kelvin) = 28,050 Joules - 32,392.6 Joules = - 4,342.6 Joules = -4.34 kJ Spontaneous

8) Determine the Go and predict whether the following reaction is spontaneous at 25oC if H° = -92.22 kJ and S° = -198.75 J/K. Determine the Go and predict whether this same reaction would be spontaneous at 500oC. N2(g) + 3 H2(g)

2 NH3(g)

Go = Ho - TSo @25o Go = - 92,220 Joules – (25 + 273 Kelvin)(-198.75 Joules/Kelvin) = - 92,220 Joules + 59,227.5 = - 32,992.5 Joules Spontaneous @500 oC Go = - 92,220 Joules – (500 + 273 Kelvin)(-198.75 Joules/Kelvin) = - 92,220 Joules + 153,633.75 Joules = 61,413.75 Joules or 61.4 kJ Not Spontaneous...