Question

The concentrations of bromides in an industrial effluent density 1.00 g/ml are determined using a bromide ion selective electrode. The 0.005 M and 0.010M standards give potential readings of -0.109 v and -0.125 v, respectively. The sample gives a reading of -0.120 v. What is the bromide concentration of the sample in ppm? Br: 79.9 g/mol

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Answer to a math question The concentrations of bromides in an industrial effluent density 1.00 g/ml are determined using a bromide ion selective electrode. The 0.005 M and 0.010M standards give potential readings of -0.109 v and -0.125 v, respectively. The sample gives a reading of -0.120 v. What is the bromide concentration of the sample in ppm? Br: 79.9 g/mol

Expert avatar
Jett
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94 Answers
Los potenciales eléctricos siguen la ecuación de Nernst. Vamos a encontrar la pendiente y la intersección para la línea de calibración primero, y luego usar eso para encontrar la concentración de la muestra.

1. Primeramente, convertimos las concentraciones de Molaridad (M) de las soluciones estándar a su logaritmo en base 10:
\log(0.005) = \log 10^{-2.3} = -2.301
\log(0.010) = \log 10^{-2} = -2

2. Usamos las lecturas de potencial para encontrar la pendiente y la intersección mediante una regresión lineal simple entre el potencial (E) y el logaritmo de la concentración (\(\log [Br^-]\)):
m = \frac{E_2 - E_1}{\log [Br^-]_2 - \log [Br^-]_1} = \frac{-0.125 - (-0.109)}{-2 - (-2.301)}
m = \frac{-0.125 + 0.109}{-2 + 2.301}
m = \frac{-0.016}{0.301}
m \approx -0.05315 \, \text{V}

3. Ahora encontramos la intersección (b) usando una de las parejas de datos:
E = m \log([Br^-]) + b
-0.109 = -0.05315(-2.301) + b
b = -0.109 + 0.1224
b = 0.0134 \, \text{V}

4. Ahora usamos la ecuación de Nernst con la lectura de la muestra para encontrar \(\log[Br^-]\):
E = m \log([Br^-]) + b
-0.120 = -0.05315 \log([Br^-]) + 0.0134
-0.120 - 0.0134 = -0.05315 \log([Br^-])
-0.1334 = -0.05315 \log([Br^-])
\log([Br^-]) = \frac{-0.1334}{-0.05315}
\log([Br^-]) \approx 2.51

5. Deshacemos el logaritmo para encontrar la concentración en M:
[Br^-] = 10^{2.51}
[Br^-] \approx 0.00309 \, \text{M}

6. Convertimos la concentración de M a ppm usando la masa molar de Br:
\text{ppm} = [Br^-] (\text{M}) \times \text{masa molar de Br} \times \frac{\text{densidad de la muestra}}{\text{densidad del agua}} \times 1000
\text{ppm} = 0.00309 \times 79.9 \times 1 \times 1000
\text{ppm} \approx 49.9 \ \text{ppm}

Respuesta: 49.9 \ \text{ppm}

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