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Montano1993 [528]
3 years ago
8

Why are the trends and exceptions to the trends in ionization energy observed? Check all that apply. Ionization energy tends to

increase down a group because the electrons get farther away from the nucleus. Ionization energy tends to increase across a period because the nuclear charge increases. Ionization energy tends to increase across a period because electrons are added to the same main energy level. The ionization energies of the elements in Group 16 tend to be slightly smaller than the elements in Group 15 because the fourth electron is added to an unfilled p orbital. The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron in the p orbital from the nucleus.
Chemistry
2 answers:
NNADVOKAT [17]3 years ago
5 0
The trends and exceptions to the trends in ionization energy observed includes;
B, ionization energy tends to increase across a period because the nuclear charge increases.
C, ionization energy tends to increase across a period because electrons are added to the same main energy level.
E, The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron, in the p orbital from the nucleus. 
Ionization energies measure the tendency of a neutral atom to resist the loss of electrons. It takes a considerable amount of energy, for example to remove an electron from a neutral fluorine atom to form a positively charged ion. <span />
Georgia [21]3 years ago
4 0

answers are

2) Ionization energy tends to increase across a period because the nuclear charge increases.

3)Ionization energy tends to increase across a period because electrons are added to the same main energy level.

5) The ionization energies of elements in Group 13 tend to be lower than the elements in Group 2 because the full s orbital shields the electron in the p orbital from the nucleus.


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A 25.0 mL solution of 0.100 M CH3COOH is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of
zaharov [31]

Answer:

a) pH = 2.88

b) pH = 4.598

c) pH = 5.503

d) pH = 8.788

e) pH = 12.097

Explanation:

  • CH3COOH ↔ CH3COO-  +  H3O+

∴ Ka = 1.75 E-5 = [H3O+]*[CH3COO-] / [CH3COOH]

a) 0.0 mL KOH:

mass balance:

⇒ <em>C</em> CH3COOH = [CH3COOH] + [CH3COO-] = 0.100 M

charge balance:

⇒ [H3O+] = [CH3COO-]

⇒ 1.75 E-5 = [H3O+]²/(0.100 - [H3O+])

⇒ [H3O+]² + 1.75 E-5[H3O+] - 1.75 E-6 = 0

⇒ [H3O+] = 1.314 E.3 M

∴ pH = - Log [H3O+]

⇒ pH = 2.88

b) 5.0 mL KOH:

  • CH3COOH + KOH ↔ CH3COONa + H2O

∴ <em>C </em>CH3COOH = ((0.025)(0.100) - (5 E-3)(0.200))/(0.025+5 E-3)

⇒ <em>C</em> CH3COOH = 0.05 M

∴ <em>C</em> KOH = ((5 E-3)(0,200))/(0.025+5 E-3) = 0.033 M

mass balance:

⇒ <em>C</em> CH3COOH + <em>C</em> KOH = [CH3COOH] + [CH3COO-] = 0.05 + 0.033 = 0.083 M

charge balance:

⇒ [H3O+] + [K+] = [CH3COO-]

⇒ [CH3COO-] = [H3O+] + 0.033

⇒ 1.75 E-5 = ([H3O+]*([H3O+] + 0.033))/(0.083 - ([H3O+] + 0.033))

⇒ 1.75 E-3 = ([H3O+]² + 0.033[H3O+])/(0.05 - [H3O+])

⇒ 8.75 E-7 - 1.75 E-5[H3O+] = [H3O+]² + 0.033[H3O+]

⇒ [H3O+]² +0.03302[H3O+] - 8.75 E-7 = 0

⇒ [H3O+] = 2.523 E-5 M

⇒ pH = 4.598

equivalent point:

  • (<em>C</em>*V)acid = (<em>C</em>*V)base

⇒ (0.100 M)*(0.025 L) = (0.200 M)( Vbase)

⇒ Vbase = 0.0125L = 12.5 mL

c) 10.0 mL KOH:

∴ <em>C</em> CH3COOH = 0.0143 M

∴ <em>C</em> KOH =  0.057 M

as in the previous point, starting from the mass and charge balances, we obtain:

⇒ [H3O+] = 3.1386 E-6 M

⇒ pH = 5.503

d) 12.5 mL KOH:

at the equivalence point, there is complete salt formation, then the pH is calculated through the salt:

  • CH3COO- + H2O ↔ CH3COOH - OH-

∴ Kw/Ka = 1 E-14/1.75 E-5 = 5.714 E-10 = [CH3COOH]*[OH-]/[CH3COO-]

∴ [CH3COO-] = (0.025)(0.100))/(0.025+0.0125) = 0.066 M

mass balance:

⇒ 0.066 = [CH3COOH] + [CH3COO-]..........(1)

charge balance:

⇒ [K+] = [OH-] + [CH3COO-] = 0.066 M.........(2)

∴ [K+] = <em>C</em> CH3COO- = 0.066 M

(1) = (2):

⇒ [OH-] = [CH3COOH].......(3)

⇒ 5.714 E-10 = [OH-]² / (0.066 - [OH-])

⇒ [OH-]² + 5.714 E-10[OH-] - 3.7712 E-11 = 0

⇒ [OH-] = 6.1408 e-6 m

⇒ pOH = 5.212

⇒ pH = 14 - pOH = 8.788

d) 15.0 mL KOH:

after the equivalence point there is salt and excess base (OH-); ph is calculated from excess base:

⇒ <em>C</em> KOH = ((0.015)(0.200) - (0.025)(0.100)) / (0.025 + 0.015) = 0.0125 M

⇒ [OH-] ≅ <em>C</em> KOH = 0.0125 M

⇒ pOH = 1.903

⇒ pH = 12.097

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