10 points. Thank you in advance?

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:

⇒ C 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

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

⇒ C CH3COOH = 0.05 M

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

mass balance:

⇒ C CH3COOH + C 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:

(C*V)acid = (C*V)base

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

⇒ Vbase = 0.0125L = 12.5 mL

c) 10.0 mL KOH:

∴ C CH3COOH = 0.0143 M

∴ C 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+] = C 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:

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

⇒ [OH-] ≅ C KOH = 0.0125 M

⇒ pOH = 1.903

⇒ pH = 12.097

8 0

3 years ago

3 H2 (g) + N2 (g) 2 NH3 (g)

ale4655 [162]

Answer:

Mass = 0.697 g

Explanation:

Given data:

Volume of hydrogen = 1.36 L

Mass of ammonia produced = ?

Temperature = standard = 273.15 K

Pressure = standard = 1 atm

Solution:

Chemical equation:

3H₂ + N₂ → 2NH₃

First of all we will calculate the number of moles of hydrogen:

PV = nRT

R = general gas constant = 0.0821 atm.L/mol.K

1atm ×1.36 L = n × 0.0821 atm.L/mol.K × 273.15 K

1.36 atm.L = n × 22.43 atm.L/mol

n = 1.36 atm.L / 22.43 atm.L/mol

n = 0.061 mol

Now we will compare the moles of hydrogen and ammonia:

H₂ : NH₃

3 : 2

0.061 : 2/3×0.061 = 0.041

Mass of ammonia:

Mass = number of moles × molar mass

Mass = 0.041 mol × 17 g/mol

Mass = 0.697 g

4 0

3 years ago

When a piece of metal is irradiated with UV radiation (λ = 162 nm), electrons are ejected with a kinetic energy of 3.54×10-19 J.

dsp73

We have that the work function of the metal

$\phi=1.227*10^{-18}J$

From the Question we are told that

UV radiation (λ = 162 nm)

Kinetic energy $K.E =3.54*10-19 J.$

Generally the equation for Kinetic energy is mathematically given as

$KE =\frac{hc}{\pi-\phi} \\\\\phi =\frac{ 6.626*10^{-34} * 3*10^8}{162*10^{-9} -3.54*10^{-19}}$

$\phi=1.227*10^{-18}J$

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8 0

3 years ago

Which of the following summertime activities consists solely of physical change?

Evgen [1.6K]

A: making s sandcastle. This is because water and sand is only a mixture, so they do not react with each other. All the rest include chemical reactions!

5 0

3 years ago

Aqueous solutions of isopropyl alcohol are commonly sold as rubbing alcohol. The boiling point of isopropyl alcohol is 82.4 °C.

Monica [59]

Answer:

This is due to more hydrogen bonding in ethylene glycol than it is in isopropyl alcohol

Explanation:

The boiling point of isopropyl alcohol is 82.4 °C it contains only a single OH group, hence intermolecular hydrogen bonding is solely responsible for it's boiling point, whereas Ethylene glycol (CH2OHCH2OH) contains 2-OH group and both intermolecular and intramolecular hydrogen bonding are responsible for the higher boiling point of ethylene glycol at 198 °C.

8 0

3 years ago