##([“B”])/([“BH”^(+)]) = 0.0281##

There are two ways in which you can approach this problem, one using the base dissociation constant and the solution’s ##pOH##, and the other one using the acid dissociation constant and the solution’s .

I’ll show you how to solve it using ##K_b## and ##[“OH”^(-)]##, and you try the other approach as practice. So, you know that you’re dealing with a weak base that has the base dissociation constant, ##K_b##, equal to ##8.91 * 10^(-6)##.

The equilibrium dissociation of the weak base, which I’ll call ##”B”## for simplicity, looks like this

##”B”_text((aq]) + “H”_2″O”_text((l]) rightleftharpoons “BH”_text((aq])^(+) + “OH”_text((aq])^(-)##

By definition, the base dissciation constant is equal to

##K_b = ([“BH”^(+)] * [“OH”^(-)])/([B])##

Use the blood’s pH to determine what the ##pOH## is

##pH_”sol” + pOH = 14 implies pOH = 14 – pH_”sol”##

##pOH = 14 – 7.4 = 6.6##

The concentration of the hydroxide ions present in solution will be

##[“OH”^(-)] = 10^(-pOH)##

##[“OH”^(-)] = 10^(-6.6) = 2.5 * 10^(-7)”M”##

Rearrange the equation for ##K_b## to get

##K_b * [“B”] = [“BH”^(+)] * [“OH”^(-)]##

##([“B”])/([“BH”^(+)]) = ([“OH”^(-)])/K_b = (2.5 * 10^(-7))/(8.91 * 10^(-6)) = color(green)(0.0281)##