Q5 TY Find [Hg22+] in equilibrium with... [FREE SOLUTION]

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Chapter 8: Q5 TY (page 168)

Find $[{Hg}_{2}^{2 +}]$ in equilibrium with 0.010MKCl saturated with ${Hg}_{2} {Cl}_{2}$ .

Short Answer

Expert verified

The value of $[{Hg}_{2}^{2 +}]$ in equilibrium with 0.010M KCI saturated ${Hg}_{2} {Cl}_{2}$ is $2.2 脳 10^{- 14} M$

Step by step solution

Concept used.

Solubility product $(K_{s p})$ :

The product of a substance's dissolved ion concentrations raised to the power of their stoichiometric coefficients is used to calculate the solubility product.

Step 2: Calculate the value of [Hg22+].

Assume that ${Hg}_{2} {Cl}_{2}$ each contribute a negligible amount of C $l^{-}$ to 0.010MKCl. This is due to the low solubility of ${Hg}_{2} {Cl}_{2}$ .

From the table 8-1 and given information the required values are,

$渭 = 0.010 M [C l^{-}] = 0.010 M 纬_{H g^{2 +}} = 0.660 a n d 纬_{C r} = 0.899$

Calculate the value of $[{Hg}_{2}^{2 +}]$ using the equation for the solubility product

$K_{sp} = 1.2 脳 10^{- 18} = [{Hg}_{2}^{2 +}] 纬_{\lg^{2 +}} {[{Cl}^{-}]}^{2} 纬^{2} {Cr}^{2} = [{Hg}_{2}^{2 +}] (0.660) {(0.01)}^{2} {(0.899)}^{2} [{Hg}_{2}^{2 +}] = 2.2 脳 10^{- 14} M$

The value of $[{Hg}_{2}^{2 +}]$ in equilibrium with 0.010M KCI saturated ${Hg}_{2} {Cl}_{2}$ is role="math" localid="1663411773731" $2.2 脳 10^{- 14} M$

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91影视

Find $[{Hg}_{2}^{2 +}]$ in equilibrium with 0.010MKCl saturated with ${Hg}_{2} {Cl}_{2}$ .

Short Answer

Step by step solution

Concept used.

Step 2: Calculate the value of [Hg22+].

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91影视

Find [Hg22+] in equilibrium with 0.010MKCl saturated withHg2Cl2.

Short Answer

Step by step solution

Concept used.

Step 2: Calculate the value of [Hg22+].

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Chemical Analysis

Nuclear Chemistry

Organic Chemistry

Chemistry Branches

Inorganic Chemistry

Making Measurements

Study anywhere. Anytime. Across all devices.

Find $[{Hg}_{2}^{2 +}]$ in equilibrium with 0.010MKCl saturated with ${Hg}_{2} {Cl}_{2}$ .