Chemistry, 10.03.2020 22:29 WhiteWinterRose

PbI2(s)⇄Pb2+(aq)+2I−(aq) Ksp=7×10−9

The dissolution of PbI2(s) is represented above.

(a) Write a mathematical expression that can be used to determine the value of S , the molar solubility of PbI2(s) .

(b) If PbI2(s) is dissolved in 1.0MNaI(aq), is the maximum possible concentration of Pb2+(aq) in the solution greater than, less than, or equal to the concentration of Pb2+(aq) in the solution in part (a) ? Explain.

Compound Ksp
PbCl2 2×10−5
PbI2 7×10−9
Pb(IO3)2 3×10−13
c) A table showing Ksp values for several lead compounds is given above. A saturated solution of which of the compounds has the greatest molar concentration of Pb2+(aq) ? Explain.

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Answer from: jasminejbronner

The mixing of the two-component is called solutions. The two components are as follow:-

Solute Solvent

According to the question. These are the answer to the following:-

The solution to the first question is;

$PbI_2(s) -- Pb^{2+}(aq) + 2I^-(aq) \\Ksp = 7 * 10^{-9}$

$K_s_p = [PB][I]^{2+}\\= s*2s\\s^3 = \frac{k_s_p}{4}$

$s= \sqrt[3]{\frac{k_s_p}{4}}$

The solution to the second question is

$s= \sqrt[3]{\frac{k_s_p}{4}}\\\\s= \sqrt[3]{\frac{7*10^-^9}{4}}$

After solving it the value of s is 1.2 *10^-^3 the maximum possible concentration of Pb²⁺ in the solution is always less than that in the solution in part (a). That is because of the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

The solution to the third question is

The molar concentration of the following is $K_s_p = [Pb{^2^+}][X^-]^2$

According to the solubility, the solution which has larger ksp will have larger solubility.

For more information, refer to the link:-

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Answer from: amourjenny

(a) $s =\sqrt [3]{\dfrac{K_{sp}}{4}}$

(b) Less than; common ion effect

Explanation:

(a) Mathematical expression

Let s = the solubility of PbI₂.

The equation for the equilibrium is

PbI₂(s) ⇌ Pb²⁺(aq) + 2I⁻(aq); Ksp = 7 × 10⁻⁹

[Pb²⁺] = s

Then [I⁻] = 2s

$K_{sp} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (2s)^{2} = 4s^{3}\\s^{3} = \dfrac{K_{sp}}{4}\\\\s =\mathbf{ \sqrt [3]{\dfrac{K_{sp}}{4}}}\\\\\text{A mathematical expression you could use is }\mathbf{s =\sqrt [3]{\dfrac{K_{sp}}{4}}}$

(b) Concentration of Pb²⁺

(i) In water

$\begin{array}{rcl}\\s &=&\sqrt [3]{\dfrac{K_{\text{sp}}}{4}}\\\\s &=&\sqrt [3]{\dfrac{7 \times 10^{-9}}{4}}\\\\s &=&\sqrt [3]{1.75 \times 10^{-9}}\\\\&=& \mathbf{1.21 \times 10^{-3}} \textbf{ mol/L}\\\end{array}$

(ii) In 1.0 mol·L⁻¹ NaI

$\begin{array}{lrccc}\rm PbI_{2} & \rightleftharpoons &\text{Pb}^{2+} & + & \text{2I}^{-} \\ & & 0 & & 1.0 \\ & & +x & & +2x \\ & & x & & 1.0+2x \\\end{array}$

$K_{\text{sp}} =\text{[Pb$^{2+}$][I$^{-}$]}^{2} = s\times (1.0 + 2s)^{2} = 7 \times 10^{-9}\\s = \mathbf{7 \times 10^{-9}} \textbf{ mol/L}$

The maximum possible concentration of Pb²⁺ in the solution is less than that in the solution in part (a).

That is because of the common ion effect. The added iodide ion forces the position of equilibrium to shift to the left, reducing the concentration of Pb²⁺.

Each compound has the same general formula, PbX₂, so they all have a similar Ksp expression,

$K_{\text{sp}} =\text{[Pb$^{2+}$][X$^{-}$]}^{2}\\\\s = \text{[Pb$^{2+}$]} = \sqrt [3]{\dfrac{K_{\text{sp}}}{4}}$

The solute with the largest Ksp value has the greatest solubility.

A saturated solution of PbCl₂ has the greatest concentration of Pb²⁺.

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did anyone actually do this? like if anyone has submitted this i rly need .

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PbI2(s)⇄Pb2+(aq)+2I−(aq) Ksp=7×10−9

The dissolution of PbI2(s) is represented above.

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