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olganol [36]
2 years ago
14

The p K a pKa of acetic acid is 4.76 . 4.76. The p K a pKa of trichloroacetic acid is 0.7 . 0.7. Calculate the equilibrium disso

ciation constant, K a , Ka, of each acid.
Chemistry
1 answer:
Andrei [34K]2 years ago
3 0

Answer:

Acetic acid Ka = 1.74 × 10⁻⁵

Trichloroacetic acid Ka = 2 × 10⁻¹

Explanation:

Let's consider the acid dissociation of acetic acid.

CH₃COOH(aq) ⇄ CH₃COO⁻(aq) + H⁺(aq)

The pKa of acetic acid is 4.76. The acid dissociation constant (Ka) is:

pKa = -log Ka

- pKa = log Ka

Ka = anti log (-pKa)

Ka = anti log (-4.76)

Ka = 1.74 × 10⁻⁵

Let's consider the acid dissociation of trichloroacetic acid.

CCl₃COOH(aq) ⇄ CCl₃COO⁻(aq) + H⁺(aq)

The pKa of trichloroacetic acid is 0.7. The acid dissociation constant (Ka) is:

pKa = -log Ka

- pKa = log Ka

Ka = anti log (-pKa)

Ka = anti log (-0.7)

Ka = 2 × 10⁻¹

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Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in
avanturin [10]
<span>5.5×10−2M in calcium chloride and 8.0×10−2M in magnesium nitrate.
What mass of sodium phosphate must be added to 1.5L of this solution to completely eliminate the hard water ion

1) Content of Ca (2+) ions

Calcium chloride = CaCl2

Ionization equation: CaCl2 ---> Ca (2+) + 2 Cl (-)

=> Molar ratios: 1 mol of CaCl2 : 1 mol Ca(2+) : 2 mol Cl(-)

Calculate the number of moles of CaCl2 in 1.5 liters of 5.5 * 10^-2 M solution

M = n / V => n = M*V = 5.5 * 10^ -2 M * 1.5 l = 0.0825 mol CaCl2

=> 0.0825 mol Ca(2+)

2) Number of phosphate ions needed to react with 0.0825 mol Ca(2+)

formula of phospahte ion: PO4 (3-)

molar ratio: 2PO4(3-) + 3Ca(2+) = Ca3 (PO4)2

Proportion: 2 mol PO4(3-) / 3 mol Ca(2+) = x / 0.0825 mol Ca(2+)

=> x = 0.0825 coml Ca(2+) * 2 mol PO4(3-) / 3 mol Ca(2+) = 0.055 mol PO4(3-)

3) Content of Mg(2+) ions

Ionization equation: Mg (NO3)2 ----> Mg(2+) + 2 NO3 (-)

Molar ratios: 1 mol Mg(NO3)2 : 1 mol Mg(2+) + 2 mol NO3(-)

number of moles of Mg(NO3)2 in 1.5 liter of 8.0 * 10^-2 M solution

n = M * V = 8.0 * 10^ -2 M * 1.5 liter = 0.12 moles Mg(NO3)2

ions of Mg(2+) = 0.12 mol Mg(NO3)2 * 1 mol Mg(2+) / mol Mg(NO3)2 = 0.12 mol Mg(2+)

4) Number of phosphate ions needed to react with 0.12 mol Mg(2+)

2PO4(3-) + 3Mg(2+) = Mg3(PO4)2

=> 2 mol PO4(3-) / 3 mol Mg(2+) = x / 0.12 mol Mg(2+)

=> x = 0.12 * 2/3 mol PO4(3-) = 0.16 mol PO4(3-)

5) Total number of moles of PO4(3-)

0.055 mol + 0.16 mol = 0.215 mol

6) Sodium phosphate

Sodium phosphate = Na3(PO4)

Na3PO4 ---> 3Na(+) + PO4(3-)

=> 1 mol Na3PO4 : 1 mol PO4(3-)

=> 0.215 mol PO4(3-) : 0.215 mol Na3PO4

mass in grams = number of moles * molar mass

molar mass of Na3 PO4 = 3*23 g/mol + 31 g/mol + 4*16 g/mol = 164 g/mol

=> mass in grams = 0.215 mol * 164 g/mol = 35.26 g

Answer: 35.26 g of sodium phosphate
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5 0
3 years ago
At a certain temperature the vapor pressure of pure methanol is measured to be . Suppose a solution is prepared by mixing of met
CaHeK987 [17]

Answer:

Partial pressure is 0.13 atm

Explanation:

CHECK THE COMPLETE QUESTION BELOW :

At a certain temperature, the vapor pressure of pure methanol is measured to be 0.43atm. Suppose a solution is prepared by mixing 88.2 g of methanol and 116.g of water. Calculate the partial pressure of methanol vapor above this solution. Be sure your answer has the correct number of significant digits. Note for advanced students: you may assume the solution is ideal.

Using Raoult´s law for ideal soultions we have

P(A) = X(A) *Pº(A)

where P(A) is the partial vapor pressure pressure of methanol,

X(A) is the mole fraction of solute (methanol) in solution,

Pº(A) is the vapor pressure of pure solute

Raoult's law states that the vapor pressure of a solution is dependent on the mole fraction of a solute added to the solution.

Raoult's law can be expressed below

Psolution = ΧsolventP0solvent.

Expressing it interns of the constituents given in the question we have

P(CH₃OH) = X(CH₃OH) x Pº(CH₃OH)

To calculate the mole fraction of CH₃OH, we make use of the formula below :

X(A) = mol (A) / ntotal

Ntotal = (sum of number of moles of A )+( moles solvent)

mol (CH₃3OH) can be calculated as :: 88.2 g/ 32 g/mol = 2.76 mol of CH₃OH

mol (H₂O) can be calculated as ::116 g/ 18 g/mol = 6.44 mol

total n = (6.44 + 2.76) mol = 9.20 mol

To calculate the partial pressure the we say;

P(CH₃OH) = (2.76 mol CH + 9.20 mol) x( 0.43 atm) = 0.13 atm

Hence, the partial pressure rounded to two significant figures is 0.13 atm

8 0
3 years ago
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