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Chapter 12: Problem 132

An experimenter makes up a solution of \(0.360 \mathrm{~mol}\) \(\mathrm{Na}_{3} \mathrm{PO}_{4}, 0.100 \mathrm{~mol} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2},\) and \(0.100 \mathrm{~mol} \mathrm{AgNO}_{3}\) in \(4.000 \mathrm{~L}\) of water solution. Note any precipitations that occur, writing a balanced equation for each. Then, calculate the molarities of each ion in the solution.

Short Answer

Expert verified
Ag鈧働O鈧 and Ca鈧(PO鈧)鈧 precipitate. Final ion M concentrations: Ag鈦 = 0, PO鈧劼斥伝 = 0.005, Ca虏鈦 = 0.025, Na鈦 = 0.270, NO鈧冣伝 = 0.050.

Step by step solution

01

Identify Precipitation Reactions

To determine if precipitations occur, we need to identify the possible ionic reactions. 1. From the compounds provided: - Na鈧働O鈧 dissociates into 3 Na鈦 and PO鈧劼斥伝 ions. - Ca(NO鈧)鈧 dissociates into Ca虏鈦 and 2 NO鈧冣伝 ions. - AgNO鈧 dissociates into Ag鈦 and NO鈧冣伝 ions.2. The potential insoluble precipitates according to solubility rules include: - Ag鈧働O鈧, which is insoluble. - Ca鈧(PO鈧)鈧, which is also insoluble.3. Thus, the precipitations that could occur are: \[ 3 ext{Ag}^+ + ext{PO}_4^{3-} ightarrow ext{Ag}_3 ext{PO}_4(s) \] \[ 3 ext{Ca}^{2+} + 2 ext{PO}_4^{3-} ightarrow ext{Ca}_3( ext{PO}_4)_2(s) \]
02

Calculate Ion Concentrations

To find molarities, use the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in Liters}} \]1. Calculate the initial molarities before any precipitation: - For Na鈦: Total moles from Na鈧働O鈧 = 0.360 mol. Molarity = \( \frac{3 imes 0.360}{4} = 0.270 \) M - For PO鈧劼斥伝: Total moles from Na鈧働O鈧 = 0.360 mol. Molarity = \( \frac{0.360}{4} = 0.090 \) M - For Ca虏鈦: Moles from Ca(NO鈧)鈧 = 0.100 mol. Molarity = \( \frac{0.100}{4} = 0.025 \) M - For Ag鈦: Moles from AgNO鈧 = 0.100 mol. Molarity = \( \frac{0.100}{4} = 0.025 \) M - For NO鈧冣伝: Total from Ca(NO鈧)鈧 and AgNO鈧 = 0.200 mol. Molarity = \( \frac{0.200}{4} = 0.050 \) M2. Adjust for precipitates: - Ag鈧働O鈧 precipitation: Ag鈦 moles = 0.075 mol from 0.025 M * 3 parts for 4 L, PO鈧劼斥伝 = 0.025 mol - Ca鈧(PO鈧)鈧 precipitation: Remaining PO鈧劼斥伝 moles = 0.070 - (0.025 for Ag鈧働O鈧 and 0.035) = 0.020 molFinal Molarities: - Ag鈦: 0 - PO鈧劼斥伝: 0.020/4 = 0.005 M - Ca虏鈦: 0.060 - 0.035 for precipitate unchanged = 0.025 M - Na鈦: 0.270 M - NO鈧冣伝: 0.050 M

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionic Equations
Understanding ionic equations can initially seem tricky, but they are just a way of representing certain chemical reactions. In any solution where ions are present, if two ions combine to form a compound that is insoluble in water, they will form a precipitate. In our case, when solutions of Na鈧働O鈧, Ca(NO鈧)鈧, and AgNO鈧 are mixed, the ions mix freely. However, according to solubility rules, Ag鈧働O鈧 and Ca鈧(PO鈧)鈧 are insoluble in water, which means they form solid precipitates. This is why the ionic equations are shown as: - For silver phosphate: \[ 3\text{Ag}^+ + \text{PO}_4^{3-} \rightarrow \text{Ag}_3\text{PO}_4(s) \] - For calcium phosphate: \[ 3\text{Ca}^{2+} + 2\text{PO}_4^{3-} \rightarrow \text{Ca}_3(\text{PO}_4)_2(s) \] These balanced equations show the formation of these precipitates from their respective ions in the solution.
Molarity Calculation
Molarity provides a straightforward way to express the concentration of a solution. It is calculated by dividing the number of moles of solute by the volume of solution in liters. Here's the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in Liters}} \] Before any precipitation occurs, you calculate each ion's molarity based on the initial moles present. For example, the sodium ions (Na鈦) from Na鈧働O鈧 have a molarity of 0.270 M, and phosphate ions (PO鈧劼斥伝) have 0.090 M. Each of the calcium (Ca虏鈦) and silver (Ag鈦) ions have a molarity of 0.025 M. After some ions form precipitates, the remaining ion concentrations are adjusted to find the final molarity. Understanding these calculations helps to predict the concentration changes due to precipitation.
Solubility Rules
Solubility rules serve as a guide to predict whether a particular compound will dissolve in water. These rules are invaluable in identifying which compounds will precipitate out of solution when two ionic solutions are mixed. In our experiment, we need to apply these rules to decide if any of the substances would form a solid precipitate. For instance, - Compounds containing the phosphate group (PO鈧劼斥伝) are mostly insoluble except when they pair with alkali metals like Na鈦. - Silver phosphate (Ag鈧働O鈧) and calcium phosphate (Ca鈧(PO鈧)鈧) are considered insoluble in water. Applying these solubility rules helps us determine which ions will bond together to form a solid, ensuring we accurately write and balance our chemical equations.
Chemical Reactions
The study of chemical reactions involves understanding the processes where substances interact to form entirely new products. Examples include precipitation reactions, where two solutions combine, and a solid forms. This is different from simple mixing, as it involves breaking and forming chemical bonds. In our scenario with Na鈧働O鈧, Ca(NO鈧)鈧, and AgNO鈧, the purpose is to identify reactions that generate precipitates. When these solutions are combined, the interactions of ions lead to the formation of silver phosphate (Ag鈧働O鈧) and calcium phosphate (Ca鈧(PO鈧)鈧). Recognizing these reactions helps in predicting the result of mixing different solutions and in understanding the changes in composition within the solution.

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