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QUESTION

# Novocaine, which is used by dentists as local anaesthetic, is a weak base with Kb = 8.91 x 10-6. Blood has a pH of 7.4. What is the ratio of concentrations of novocaine to its conjugate acid in the blood stream?

(["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)