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3 years ago

14

Enter the Ksp expression for the solid AB2 in terms of the molar solubility x. AB2 has a molar solubility of 3.72×10−4 M. What i

s the value of the solubility product constant for AB2?

1 answer:

AlekseyPX3 years ago

5 0

Answer:

2.06 × 10⁻¹⁰

Explanation:

Let's consider the solution of a generic compound AB₂.

AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)

We can relate the molar solubility (S) with the solubility product constant (Kps) using an ICE chart.

AB₂(s) ⇄ A²⁺(aq) + 2B⁻(aq)

I 0 0

C +S +2S

E S 2S

The solubility product constant is:

Kps = [A²⁺] × [B⁻]² = S × (2S)² = 4 × S³ = 4 × (3.72 × 10⁻⁴)³ = 2.06 × 10⁻¹⁰

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Consider a closed containing a solid in equilibrium with its vapor. The volume of the solid is much less than that of the contai

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2 years ago

What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig

Mashutka [201]

Answer:

$P_H =2.86$

$c=1.4\times 10^{-4}$

Explanation:

first write the equilibrium equaion ,

$C_3H_6O_3$ ⇄ $C_3H_5O_3^{-} +H^{+}$

assuming degree of dissociation $\alpha$ =1/10;

and initial concentraion of $C_3H_6O_3$ =c;

At equlibrium ;

concentration of $C_3H_6O_3 = c-c\alpha$

$[C_3H_5O_3^{-} ]= c\alpha$

$[H^{+}] = c\alpha$

$K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}$

$\alpha$ is very small so $1-\alpha$ can be neglected

and equation is;

$K_a = {c\alpha \times \alpha}$

$[H^{+}] = c\alpha$ = $\frac{K_a}{\alpha}$

$P_H =- log[H^{+} ]$

$P_H =-logK_a + log\alpha$

$K_a =1.38\times10^{-4}$

$\alpha = \frac{1}{10}$

$P_H= 3.86-1$

$P_H =2.86$

composiion ;

$c=\frac{1}{\alpha} \times [H^{+}]$

$[H^{+}] =antilog(-P_H)$

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3 years ago

Calculate the molality of a solution formed by adding 6.30 g NH4CL to 15.7 g of water

babymother [125]

Answer:

Molality = 7.5 mol/kg

Explanation:

Given data:

Mass of NH₄Cl = 6.30 g

Mass of water = 15.7 g (15.7/1000 =0.016 kg)

Molality = ?

Solution:

Formula of molality:

Molality = Moles of solute / mass of solvent in gram

Now we will first calculate the number of moles of solute( NH₄Cl )

Number of moles = mass/ molar mass

Molar mass of NH₄Cl = 53.491 g/mol

Number of moles = 6.30 g/ 53.491 g/mol

Number of moles = 0.12 mol

Now we will calculate the molality.

Molality = Moles of solute / mass of solvent in gram

Molality = 0.12 mol / 0.016 kg

Molality = 7.5 m

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Molality = 7.5 mol/kg

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3 years ago

The combustion of 0.570 g of benzoic acid (ΔHcomb = 3,228 kJ/mol; MW = 122.12 g/mol) in a bomb calorimeter increased the tempera

torisob [31]

Answer:

The temperature change from the combustion of the glucose is 6.097°C.

Explanation:

Benzoic acid;

Enthaply of combustion of benzoic acid = 3,228 kJ/mol

Mass of benzoic acid = 0.570 g

Moles of benzoic acid = $\frac{0.570 g}{122.12 g/mol}=0.004667 mol$

Energy released by 0.004667 moles of benzoic acid on combustion:

$Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J$

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Change in temperature of the calorimeter = ΔT = 2.053°C

$Q=C\times \Delta T$

$15,066.8 J=C\times 2.053^oC$

$C=7,338.92 J/^oC$

Glucose:

Enthaply of combustion of glucose= 2,780 kJ/mol.

Mass of glucose=2.900 g

Moles of glucose = $\frac{2.900 g}{180.16 g/mol}=0.016097 mol$

Energy released by the 0.016097 moles of calorimeter combustion:

$Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J$

Heat capacity of the calorimeter = C (calculated above)

Change in temperature of the calorimeter on combustion of glucose = ΔT'

$Q'=C\times \Delta T'$

$44,749.1 J=7,338.92 J/^oC\times \Delta T'$

$\Delta T'=6.097^oC$

The temperature change from the combustion of the glucose is 6.097°C.

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