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Amanda [17]
3 years ago
13

Which of the halide ions (f−, cl−, br−, and i−) is the most stable base? which is the least stable base?

Chemistry
2 answers:
rusak2 [61]3 years ago
6 0

Explanation:

It is known that on moving down a group there will occur a decrease in electronegativity of non-metals. Whereas stability of their conjugate bases increases on moving down the group.

This also means that their acid strength also increases.

For example, HX + H_{2}O \rightarrow H_{3}O^{+} + X^{-}

where, X = F, Cl, Br or I)

Therefore, acidity will increase in the following order.

                      HI > HBr > HCl > HF

Hence, the stability of their conjugate bases will be as follows.

                I^{-} > Br^{-} > Cl^{-} > F^{-}

Thus, we can conclude that the most stable base is I^{-} and least stable base is F^{-}.

Nezavi [6.7K]3 years ago
3 0
You should take note that the question is about stability. A compound is stable if it does not easily react with other elements. Hence, its reactivity must be low. As you move down the group, reactivity decreases. So, the halide at the very bottom is the least reactive. It would then be logical that the most stable conjugate base is I⁻ and the least stable conjugate base is the most reactive which is F⁻.
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Answer:

12.99

Explanation:

<em>A chemist dissolves 716. mg of pure potassium hydroxide in enough water to make up 130. mL of solution. Calculate the pH of the solution. (The temperature of the solution is 25 °C.) Be sure your answer has the correct number of significant digits.</em>

Step 1: Given data

  • Mass of KOH: 716. mg (0.716 g)
  • Volume of the solution: 130. mL (0.130 L)

Step 2: Calculate the moles corresponding to 0.716 g of KOH

The molar mass of KOH is 56.11 g/mol.

0.716 g × 1 mol/56.11 g = 0.0128 mol

Step 3: Calculate the molar concentration of KOH

[KOH] = 0.0128 mol/0.130 L = 0.0985 M

Step 4: Write the ionization reaction of KOH

KOH(aq) ⇒ K⁺(aq) + OH⁻(aq)

The molar ratio of KOH to OH⁻is 1:1. Then, [OH⁻] = 0.0985 M

Step 5: Calculate the pOH

We will use the following expression.

pOH = -log [OH⁻] = -log 0.0985 = 1.01

Step 6: Calculate the pH

We will use the following expression.

pH + pOH = 14

pH = 14 - pOH = 14 -1.01 = 12.99

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To determine the empirical formula of a compound, we need to follow a series of steps.

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