Morphine is a weak base. a 0.150 m solution of morphine has a ph of 10.5. what is the kb for morphine?

Let B represent the unprotonated form of morphine, HB represent the originated form. The equation is
B + H2O <—> HB + OH-
The KB expression is [HB][OH-]/[B]. Since pH is 10.5, pOH is 3.5, and [OH-] = 10^-3.5. [HB] = [OH-] = 10^-3.5, [B] = 0.150 - [HB] = 0.150 - 10^-3.5.
Plugging in:
KB = [HB][OH-]/[B] = 10^-3.5 * 10^-3.5 / (0.150 - 10^-3.5) = 6.68 x10^-7

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ardni313 ardni313 · Answer 2 · 2019-11-05T10:00:35+01:00

The kb for morphine : [tex]Kb=6.6.10^{-7}[/tex]

Further explanation

pH is the degree of acidity of a solution that depends on the concentration of H⁺ ion. The greater the value the more acidic the solution and the smaller the pH.

pH = - log [H⁺]

pOH⁻ = 14 - pH

So that the two quantities between pH and [H⁺] are inversely proportional because they are associated with negative values.

A solution whose value is different by n has a difference in the concentration of H⁺ ion of 10ⁿ.

So we use a weak base formula to find the Kb

[tex]\displaystyle [OH^-]=\sqrt{Kb.M}[/tex]

M = weak base concentration

Given :

Morphine is a weak base with concentration 0.150 M

pH = 10.5

then

pOH = 14 - 10.5

pOH = 3.5[tex]\displaystyle [OH^-]=10^{-3.5}[/tex]

We input this to weak base formula :

[tex]\displaystyle [OH^-]=\sqrt{Kb.M}[/tex]

[tex]\displaystyle 10^{-7}=Kb.0.15\\\\Kb=\frac{10^{-7}}{0.15}\\\\Kb=6.6.10^{-7}[/tex]

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Keywords : weak base, morphine, Kb, pH,pOH