Analyzing Solids

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Household vinegar contains the organic compound acetic acid with chemical formula, $CH_{3}COOH$ . If a vinegar sample contains $0.1024\ moles$ of acetic acid, calculate the percent (mass/mass) of the $CH_{3}COOH$ in the vinegar sample.

Density of vinegar is the following:

$CH_{3}COOH=1.05\frac{g}{mL}$

Explanation

Convert the moles of $CH_{3}COOH$ to grams:

$0.1024\ moles \times \frac{60.05g}{mole}=6.1491g\ CH_{3}COOH$

To calculate the percentage of $CH_{3}COOH$ in the vinegar we need to use the following formula:

$%CH_{3}COOH=\frac{grams\ of\ CH_{3}COOH}{grams\ of\ vinegar}$

Therefore,

$\frac{6.15g}{10.00g}\times 100=61.5%\ CH_{3}COOH$

Density of vinegar is the following:

$CH_{3}COOH=1.05\frac{g}{mL}$

Explanation

Convert the moles of $CH_{3}COOH$ to grams:

$0.1024\ moles \times \frac{60.05g}{mole}=6.1491g\ CH_{3}COOH$

To calculate the percentage of $CH_{3}COOH$ in the vinegar we need to use the following formula:

$%CH_{3}COOH=\frac{grams\ of\ CH_{3}COOH}{grams\ of\ vinegar}$

Therefore,

$\frac{6.15g}{10.00g}\times 100=61.5%\ CH_{3}COOH$

Determining the molecular ion peak (parent peak) in mass spectroscopy allows you to determine what characteristic of a mystery molecule?

Molecular weight

Molecular charge

Functional groups

Nuclear charge

Explanation

The molecular ion peak is determined using mass spectrometry. The parent peak is formed when a mystery molecule does not fragment, and simply loses an electron. This means that the mass to charge ratio of this peak will allow us to determine the molecular weight of the compound.

Determining the molecular ion peak (parent peak) in mass spectroscopy allows you to determine what characteristic of a mystery molecule?

Molecular weight

Molecular charge

Functional groups

Nuclear charge

Explanation

The $K_{sp}$ of is $1.8*10^{-14}$ . What is the molar solubility of in water?

$1.3*10^{-7} \textup{ M}$

$4.3*10^{-5}\textup{ M}$

$3.1*10^{-2} \textup{ M}$

$0.011\textup{ M}$

$3.22\textup{ M}$

Explanation

The equation for the dissolution of in water is:

$CdCO_{3(s)}\rightleftharpoons Cd^{2+}_{(aq)}+CO_{3}^{2-}_{(aq)}$

The $K_{sp}$ for the above equation is:

$K_{sp}=[Cd^{2+}][CO_{3}^{2-}]=1.8 *10^{-14}$

Due to the molar ratios of the species of , the concentration of $Cd^{2+}$ and $CO_3^{2-}$ should be equal when dissolved:

$[Cd^{2+}]=[CO_{3}^{2-}]=X$

Plugging into the $K_{sp}$ equation gives:

$K_{sp}=X* X=X^{2}=1.8*10^{-14}$

$X=\sqrt{1.8* 10^{-14}}=1.3*10^{-7}$

Therefore, the solubility of in water is $1.3*10^{-7}\textup{ M}$

The $K_{sp}$ of is $1.8*10^{-14}$ . What is the molar solubility of in water?

$1.3*10^{-7} \textup{ M}$

$4.3*10^{-5}\textup{ M}$

$3.1*10^{-2} \textup{ M}$

$0.011\textup{ M}$

$3.22\textup{ M}$

Explanation

The equation for the dissolution of in water is:

$CdCO_{3(s)}\rightleftharpoons Cd^{2+}_{(aq)}+CO_{3}^{2-}_{(aq)}$

The $K_{sp}$ for the above equation is:

$K_{sp}=[Cd^{2+}][CO_{3}^{2-}]=1.8 *10^{-14}$

Due to the molar ratios of the species of , the concentration of $Cd^{2+}$ and $CO_3^{2-}$ should be equal when dissolved:

$[Cd^{2+}]=[CO_{3}^{2-}]=X$

Plugging into the $K_{sp}$ equation gives:

$K_{sp}=X* X=X^{2}=1.8*10^{-14}$

$X=\sqrt{1.8* 10^{-14}}=1.3*10^{-7}$

Therefore, the solubility of in water is $1.3*10^{-7}\textup{ M}$

For the titration of $0.001 M\ Ag^{2+}$ with solution of $Cl^-_{(aq)}$ , the end point occurs upon addition of of the $Ag^{2+}$ solution. Determine the initial concentration of chloride that was present in the original solution.

$0.0013\ M$

$1.321\ M$

$0.0026\ M$

$0.063\ M$

Explanation

The equivalence point is when enough titrant has been added so that the number of moles of titrant equals the number of moles of analyte. We must first determine the number of moles of silver ions at the equivalence point.

At the equivalence point the following occurs in solution:

$moles\ of\ Ag^{2+}=moles\ of\ Cl^{-}$

The number of moles of silver ions at equivalence point is calculated as follows:

$0.200\ L\times \frac{0.001\ moles}{L}=2.0\times 10^{-4}\ moles\ Ag^{2+}$

$Molarity=\frac{moles\ of\ solute}{Liters\ of\ solution}$

$Liters\ of\ solution= Original\ volume\ of\ Cl^{-}\ solution$

Plugging these values into the equation gives:

$\frac{2\times10^{-4}\ moles}{0.150\ L}=0.0013\ M\ Cl^{-}$

$0.0013\ M$

$1.321\ M$

$0.0026\ M$

$0.063\ M$

Explanation

At the equivalence point the following occurs in solution:

$moles\ of\ Ag^{2+}=moles\ of\ Cl^{-}$

The number of moles of silver ions at equivalence point is calculated as follows:

$0.200\ L\times \frac{0.001\ moles}{L}=2.0\times 10^{-4}\ moles\ Ag^{2+}$

$Molarity=\frac{moles\ of\ solute}{Liters\ of\ solution}$

$Liters\ of\ solution= Original\ volume\ of\ Cl^{-}\ solution$

Plugging these values into the equation gives:

$\frac{2\times10^{-4}\ moles}{0.150\ L}=0.0013\ M\ Cl^{-}$

Write the solubility product, $K_{sp}$ , for $BaSO_4_{(s)}$ .

$K_{sp} = [Ba^{2+}][SO_{4^}^{}^{2-}]$

$K_{sp} = [Ba][SO_{4}]^{2-}$

$K_{sp} = [Ba][SO_{4}]$

$K_{sp} = [Ba]^{2+}[SO_{4}]^{2-}$

$K_{sp} = [Ba]^{4+}[SO_{4}]^{2-}$

Explanation

The solubility product constant ( $K_{sp}$ ) is an expression that describes the extent to which a compound is soluble in an aqueous solution. It describes the equilibrium between a solid and its constituent ions in a solution. The equilibrium constant for $BaSO_4_{(s)}$ can be written as:

$K_{eq} = \frac{[Ba^{2+}][SO_{4^}^{}^{2-}]}{[BaSO_{4}(s)]}$

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

$K_{eq}*[BaSO_{4}]=[Ba^{2+}][SO_{4^}^{}^{2-}]$

The product, $K_{eq}*[BaSO_{4}]$ , can be considered a constant expression called the solubility product constant ( $K_{sp}$ ) and the equation can be written in the form:

$K_{sp} = [Ba^{2+}][SO_{4^}^{}^{2-}]$

Write the solubility product, $K_{sp}$ , for $BaSO_4_{(s)}$ .

$K_{sp} = [Ba^{2+}][SO_{4^}^{}^{2-}]$

$K_{sp} = [Ba][SO_{4}]^{2-}$

$K_{sp} = [Ba][SO_{4}]$

$K_{sp} = [Ba]^{2+}[SO_{4}]^{2-}$

$K_{sp} = [Ba]^{4+}[SO_{4}]^{2-}$

Explanation

$K_{eq} = \frac{[Ba^{2+}][SO_{4^}^{}^{2-}]}{[BaSO_{4}(s)]}$

The denominator represents solid barium sulfate which is considered a constant. When the equation is rearranged, it becomes:

$K_{eq}*[BaSO_{4}]=[Ba^{2+}][SO_{4^}^{}^{2-}]$

The product, $K_{eq}*[BaSO_{4}]$ , can be considered a constant expression called the solubility product constant ( $K_{sp}$ ) and the equation can be written in the form:

$K_{sp} = [Ba^{2+}][SO_{4^}^{}^{2-}]$

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