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Setler
2 months ago
15

A 5.36–g sample of NH4Cl was added to 25.0 mL of 1.00 M NaOH and the resulting solution diluted to 0.100 L. (a) What is the pH o

f this buffer solution? (b) Is the solution acidic or basic? (c) What is the pH of a solution that results when 3.00 mL of 0.034 M HCl is added to the solution?
Chemistry
1 answer:
castortr0y [3K]2 months ago
7 0
La segunda parte del procedimiento para la solución es la siguiente... 5.36g de NH₄Cl junto con 25ml de NaOH a 1M se expresa como (5.36g/53g) NH₄Cl y 0.025L de NaOH. De esto, se derivan 0.101 moles de NH₄Cl y 0.025 moles de NaOH. Así que tenemos (0.101 moles/0.025L) de NH₄Cl + (0.025 moles/0.025L) de NaOH, resultando en 4.045M de NH₄Cl y 1.000M de NaOH. Finalmente, al calcular la concentración de NH₄OH y componentes adicionales, obtenemos que 0.025M NH₄OH + 4.02M NH₄⁺ determinan la solución buffer. La pH de la solución buffer se determina como sigue: NH₄OH ⇄ NH₄⁺ + OHˉ C(i) 0.025M 4.02M 0M ΔC -x +x +x C(eq) (0.025-x)M (4.02+x)M x que se aproxima a 0.025M, 4.02M. Usando Kb, se tiene [OHˉ] = [(1.8 x 10ˉ⁵)(0.025)/(4.02)]M y se calcula pOH = -log[OHˉ] = -log(1.12 x 10ˉ⁷) resulta en 6.95, proporcionando un pH de 7.04. Luego, al añadir 3ml de 0.034M HCl, se halla que la concentración de HCl es de 3.643 x 10ˉ³M, reduciendo la concentración de OHˉ a 5.55 x 10ˉ⁸M, llevando el pOH a 7.26 y el pH a 6.74.
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