What is the pH of a 0.30 mol L^-1 pyridine solution that has K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is below: $$\mathrm { C } _ { 5 } \mathrm { H } _ { 5 } \mathrm {~N} ( a q ) + \mathrm { H } _ { 2 } \mathrm { O } ( l ) \rightleftharpoons \mathrm { C } _ { 5 } \mathrm { H } _ { 5 } \mathrm { NH } ^ { - } ( a q ) + \mathrm { OH } ^ { - } ( a q )$$ A)4.62 B)8.72 C)9.38 D)10.38

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The equilibrium reaction for the dissociation of pyridine is: C5H5N + H2O ⇌ C5H5NH- + OH- The equilibrium constant expression for this reaction is: Kb = [C5H5NH-][OH-]/[C5H5N][H2O] where Kb is the base dissociation constant. The given value K_b = 1.9 × 10^-9 is actually Kb for pyridine. To find the pOH (negative log of the hydroxide ion concentration) of the solution, we can use the expression: Kb = [C5H5NH-][OH-]/[C5H5N][H2O] Let x be the concentration of OH- ions produced in the solution. Then, the concentrations of C5H5NH- and C5H5N are equal to 0.30 mol/L (initial concentration) - x, and the concentration of H2O is essentially constant. Therefore, Kb = x^2/(0.30 - x) = 1.9 × 10^-9 Solving for x, we get: x = 5.38 × 10^-6 mol/L The pOH of the solution is: pOH = -log[OH-] = -log(5.38 × 10^-6) = 5.27 Finally, we can find the pH of the solution using the relationship: pH + pOH = 14 pH = 14 - pOH = 14 - 5.27 = 8.73 Therefore, the correct answer is C) 9.38.

What Is the PH of a 0 A)462 B)8

What Is the PH of a 0 A)462
B)8