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3 years ago

Based on the position of the elements in the periodic table, what is the charge of the ions they are most likely to form?

Chemistry

1 answer:

Rainbow [258]3 years ago

4 0

Explanation:

The charge on an ion denotes the amount of electrons lost or gained or even shared by an atom.

An atom will lose, gain or share an equal amount of electrons that will make it stable and achieve an octet and perfect configuration.

This is often synonymous with the component number of electrons in their outermost shell. The valence shell.

For metals on the periodic table, they are always willing to give up electrons due to their large electropositivity.

Metals will give up the number of electrons in their outermost shell to become stable.

Groups 1 and Group 2 will have charges +1 and +2 on them.

For non-metals , they will gain the exact number of electrons that will make them stable. We must note that half -filled and fully filled orbitals are equally stable.

Elements in groups 6 and 7 are electronegative and will have a charge of -2 and -1 respectively.

For those in groups 3, 4 and 5 they either gain, lose or share commensurate amount of electrons that will make them stable.

Group 8 elements are stable and have no charges.

Generally on the periodic table, metals are to the left and are always positively charged. Non - metals are to the right and are negatively charged.

learn more:

Periodic table brainly.com/question/2690837

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Answer: The pH of acid solution is 4.58

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$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

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Molarity of KOH solution = 1.1000 M

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Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

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Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

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