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UNO [17]
2 years ago
15

(1) The solubility of Salt AB2(S) IS 5mol/dm^3.

Chemistry
1 answer:
Svetlanka [38]2 years ago
6 0

Answer:

a. Ksp = 4s³

b. 5.53 × 10⁴ mol³/dm⁹

Explanation:

a. Obtain an expression for the solubility product of AB2(S),in terms of s.

AB₂ dissociates to give

AB₂ ⇄ A²⁺ + 2B⁻

Since 1 mole of AB₂ gives 1 mole of A and 2 moles of B, we have the mole ratio as

AB₂ ⇄ A²⁺ + 2B⁻

1 : 1 : 2

Since the solubility of AB₂ is s, then the solubility of A is s and that of B is 2s

So, we have

AB₂ ⇄ A²⁺ + 2B⁻

[s]        [s]    [2s]

So, the solubility product Ksp = [A²⁺][B⁻]²

= (s)(2s)²

= s(4s²)

= 4s³

b. Calculate the Ksp of AB₂, given that solubility is 2.4 × 10³ mol/dm³

Given that the solubility of AB is 2.4 × 10³ mol/dm³ and the solubility product Ksp = [A²⁺][B⁻]² = 4s³ where s = solubility of AB = 2.4 × 10³ mol/dm³

Substituting the value of s into the equation, we have

Ksp = 4s³

= 4(2.4 × 10³ mol/dm³)³

= 4(13.824 × 10³ mol³/dm⁹)

= 55.296 × 10³ mol³/dm⁹

= 5.5296 × 10⁴ mol³/dm⁹

≅ 5.53 × 10⁴ mol³/dm⁹

Ksp = 5.53 × 10⁴ mol³/dm⁹

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Answer:

Approximately 0.08\; \rm mol, assuming that this gas is an ideal gas.

Explanation:

Look up the standard room temperature and pressure:25\; \rm ^{\circ}C and P = 101.325 \; \rm kPa.

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\begin{aligned} T &= 25\; \rm ^{\circ}C \\ &= (25 + 273.15)\; \rm K \\ &= 293.15\; \rm K\end{aligned}.

\begin{aligned}P &= 101.325\; \rm kPa \\ &= 101.325 \; \rm kPa \times \frac{10^{3}\; \rm Pa}{1\; \rm kPa} \\ &= 1.01325 \times 10^{5}\; \rm Pa\end{aligned}.

\begin{aligned}V &= 2\; \rm dm^{3} \\ &= 2 \; \rm dm^{3} \times \frac{1\; \rm m^{3}}{10^{3}\; \rm dm^{3}} \\ &= 2 \times 10^{-3}\; \rm m^{3}\end{aligned}.

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Let n denote the number of moles of this gas in that V = 2\; \rm dm^{3}. By the ideal gas law, if this gas is an ideal gas, then the following equation would hold:

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Rearrange this equation and solve for n:

\begin{aligned}n &= \frac{P \cdot V}{R \cdot T} \\ &\approx \frac{1.01325 \times 10^{5}\; {\rm Pa} \times 2 \times 10^{-3}\; {\rm m^{3}}}{8.31 \; {\rm m^{3} \cdot Pa \cdot mol^{-1} \cdot K^{-1}} \times 293.15\; {\rm K}} \\ &\approx 0.08\; \rm mol\end{aligned}.

In other words, there is approximately 2\; \rm mol of this gas in that V = 2\; \rm dm^{3}.

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