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

The kb for morphine : 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

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

We input this to weak base formula :

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

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