Question #1809d

1 Answer
Oct 2, 2015

Answer:

#K_b = 3.7 * 10^(-6)#

Explanation:

Start by using the known pH of the solution to find the molarity of the hydrogen ions, #"H"^(+)#. This will help you find the acid dissociation constant, #K_a#, for quinine's conjugae acid, #"QH"^(+)#.

So, you know that quinine's conjugate acid, #"QH"^(+)#, will ionize in aqueous solution to give

#"OH"_text((aq])^(+) -> "Q"_text((aq]) + "H"_text((aq])^(+)#

The concentration of the conjugate acid is simply the ratio between the number of moles you have and the volume of the solution

#["QH"^(+)] = "0.23 moles"/"1.0 L" = "0.23 M"#

The concentration of #"H"^(+)# will be

#["H"^(+)] = 10^(-"pH") = 10^(-4.58) = 2.63 * 10^(-5)"M"#

Since the dissociation reactrion produces equal numbers of moles of #"Q"# and #"H"^(+)#, you get that

#["Q"] = ["H"^(+)] = 2.63 * 10^(-5)"M"#

The acid dissociation constant for this equilibrium will thus be

#K_a = (["Q"] * ["H"^(+)])/(["QH"^(+)])#

#K_a = (2.63 * 10^(-5) * 2.63 * 10^(-5))/0.23 = 2.7 * 10^(-9)#

Finally, to get the base dissociation constant, #K_b#, use the fact that

#K_a * K_b = K_W" "#, where

#K_W# - the water self-ionization constant, equal to #10^(-14)#.

This means that #K_b# will be equal to

#K_b = K_W/K_a = 10^(-14)/(2.7 * 10^(-9)) = color(green)(3.7 * 10^(-6))#