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Codeine (C18H21NO3) is a prescription medicine used to treat mild to moderate pain in adults. It is also a weak base with a pKb of 6.05. Although commonly prescribed as a tablet, codeine is also available as an oral solution containing up to 20 mg of codeine per mL which allows doses to be highly controlled by medical professionals.Given that the molar mass of codeine is 299.36 g mol-1, calculate the pH of a 12 mL solution containing 189 mg of codeine.

Question

Codeine (C18H21NO3) is a prescription medicine used to treat mild to moderate pain in adults. It is also a weak base with a pKb of 6.05. Although commonly prescribed as a tablet, codeine is also available as an oral solution containing up to 20 mg of codeine per mL which allows doses to be highly controlled by medical professionals.Given that the molar mass of codeine is 299.36 g mol-1, calculate the pH of a 12 mL solution containing 189 mg of codeine.

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Solution

Para calcular el pH de una solución de codeína, primero necesitamos determinar la concentración molar de la codeína en la solución. Luego, utilizaremos la constante de ionización de la base (Kb) para encontrar la concentración de iones hidroxilo (OH-) y finalmente calcular el pH.

  1. Calcular la concentración molar de codeína en la solución:

    • Masa de codeína: 189 mg = 0.189 g
    • Volumen de la solución: 12 mL = 0.012 L
    • Molaridad (M) = moles de soluto / litros de solución

    Primero, convertimos la masa de codeína a moles: Moles de codeıˊna=MasaMasa molar=0.189 g299.36 g/mol6.31×104 moles \text{Moles de codeína} = \frac{\text{Masa}}{\text{Masa molar}} = \frac{0.189 \text{ g}}{299.36 \text{ g/mol}} \approx 6.31 \times 10^{-4} \text{ moles}

    Ahora, calculamos la molaridad: Molaridad=6.31×104 moles0.012 L0.0526 M \text{Molaridad} = \frac{6.31 \times 10^{-4} \text{ moles}}{0.012 \text{ L}} \approx 0.0526 \text{ M}

  2. Calcular la constante de ionización de la base (Kb):

    Dado que pKb = 6.05, podemos encontrar Kb usando la relación: Kb=10pKb=106.058.91×107 \text{Kb} = 10^{-\text{pKb}} = 10^{-6.05} \approx 8.91 \times 10^{-7}

  3. Determinar la concentración de iones OH-:

    Para una base débil, la concentración de iones OH- se puede aproximar usando la expresión: Kb=[OH]2[Base inicial] \text{Kb} = \frac{[\text{OH}^-]^2}{[\text{Base inicial}]}

    Donde [Base inicial] es la concentración de codeína, que es 0.0526 M. Entonces: 8.91×107=[OH]20.0526 8.91 \times 10^{-7} = \frac{[\text{OH}^-]^2}{0.0526}

    Resolviendo para [OH-]: [OH]2=8.91×107×0.05264.69×108 [\text{OH}^-]^2 = 8.91 \times 10^{-7} \times 0.0526 \approx 4.69 \times 10^{-8} [OH]4.69×1086.85×104 M [\text{OH}^-] \approx \sqrt{4.69 \times 10^{-8}} \approx 6.85 \times 10^{-4} \text{ M}

  4. Calcular el pOH y luego el pH:

    pOH=log[OH]=log(6.85×104)3.16 \text{pOH} = -\log [\text{OH}^-] = -\log (6.85 \times 10^{-4}) \approx 3.16

    Finalmente, usando la relación entre pH y pOH: pH=14pOH=143.16=10.84 \text{pH} = 14 - \text{pOH} = 14 - 3.16 = 10.84

Por lo tanto, el pH de la solución de codeína es aproximadamente 10.84.

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