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A 5.36–g sample of NH₄Cl was added to 25.0 mL of 1.00 M NaOH and the resulting solution Calculate the final pH.

User Olasimbo
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Final answer:

The question involves calculating the pH of a solution where NH4Cl reacts with NaOH and is then diluted. However, without additional details such as the reaction volume after dilution or the dissociation constant (Ka) for NH₄+, the exact pH cannot be determined directly from the information provided.

Step-by-step explanation:

The question is asking how to calculate the final pH of a solution when a 5.36-g sample of NH₄Cl was added to 25.0 mL of 1.00 M NaOH and the resulting solution was diluted to 0.100 L. Ammonium chloride (NH₄Cl) reacts with sodium hydroxide (NaOH) to form ammonia, water, and sodium chloride (NaCl).

As ammonia is a weak base, it only partially dissociates in water. The concentration of hydroxide ions (OH-) can be calculated, which can then be used to find the pOH, and subsequently the pH of the solution, using the formula pH = 14.00 - pOH.

The given information lacks the full detail to calculate the pH directly. But based on general knowledge, when NH4Cl reacts with NaOH, the following reaction occurs: NH₄Cl + NaOH → NH₃ + H₂O + NaCl.

Also, at the equivalence point in a titration, the pH is 7.00 if a strong acid is titrated with a strong base; however, if a weak base or acid is involved, the pH at equivalence can be different from 7. This can help predict the nature of the resulting solution after the reaction.

User Modesty
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