Question

Propose the experimental design to determine the sodium hydroxide concentration (NaOH), the sample has a total volume of 150 mL. For the titration, Nitric Acid (HNO3). Volumetric flasks of 50 mL, 100 mL and 250 mL are available. According to the technicians' notes, the titration is carried out with 2 M titrant. a) Prepare the titrant at 2 M b) Determine the concentration of the analyte if the change is at 0.11 L

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Answer to a math question Propose the experimental design to determine the sodium hydroxide concentration (NaOH), the sample has a total volume of 150 mL. For the titration, Nitric Acid (HNO3). Volumetric flasks of 50 mL, 100 mL and 250 mL are available. According to the technicians' notes, the titration is carried out with 2 M titrant. a) Prepare the titrant at 2 M b) Determine the concentration of the analyte if the change is at 0.11 L

Expert avatar
Cristian
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111 Answers
a) \text{Preparar el valorante a 2 M}

Para preparar una disolución de valorante de HNO3 (Ácido Nítrico) a una concentración de 2 M, primero es necesario entender el volumen del matraz que se seleccionará para diluir HNO3.

Para un volumen de 100 mL:
C_{HNO3} = 2 \, \text{M}
V_{HNO3} = 0.10 \, \text{L}
Por lo tanto, la cantidad de moles de HNO3 necesaria es:
\text{moles de } HNO3 = C_{HNO3} \times V_{HNO3}
= 2 \, \text{M} \times 0.10 \, \text{L}
= 0.20 \, \text{moles} .

b) \text{Determinar la concentración del analito si el viraje es a 0.11 L}

- Se asume que hay una titulación 1:1 entre HNO3 y NaOH.

Moles de HNO3 usadas:
C_{HNO3} = 2 \, \text{M}
V_{HNO3 (utilizado)} = 0.11 \, \text{L}
\text{moles de } HNO3 = C_{HNO3} \times V_{HNO3 (utilizado)}
= 2 \, \text{M} \times 0.11 \, \text{L}
= 0.22 \, \text{moles}

- Como la reacción es 1:1, los moles de NaOH serán iguales:
\text{moles de } NaOH = 0.22 \, \text{moles}

- Calculamos la concentración de NaOH:
V_{NaOH} = 0.150 \, \text{L}
C_{NaOH} = \frac{\text{moles de } NaOH}{V_{NaOH}}
= \frac{0.22 \, \text{moles}}{0.150 \, \text{L}}
= 1.467 \, \text{M}

Por lo tanto,
\text{Concentración de NaOH} = 1.467 \, \text{M} .

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