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Write molecular and net ionic equations for the successive neutralizations of each acidic hydrogen of phosphoric acid by calcium hydroxide solution. \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is insoluble; assume that the acid salts are soluble.

Short Answer

Expert verified
The molecular equations for the successive neutralizations involve the formation of calcium phosphate and calcium monohydrogen phosphate; net ionic equations exclude duplicate ions.

Step by step solution

01

Write the Balanced Molecular Equation for the First Neutralization

Observe that phosphoric acid (H₃PO₄) can donate three hydrogen ions, and initially, we will deal with the first ion. Calcium hydroxide (Ca(OH)₂) reacts with phosphoric acid:\[\text{2 } \mathrm{H_3PO_4} + \text{3 } \mathrm{Ca(OH)_2} \rightarrow \text{6 } \mathrm{H_2O} + \text{Ca_3(PO_4)_2} \]
02

Write the Balanced Molecular Equation for the Second Neutralization

Next, we proceed to the second neutralization where the molecule now has two acidic hydrogens to donate:\[\text{HPO}_4^{2-} + \text{Ca(OH)_2} \rightarrow \text{3 } \mathrm{Ca^{2+} + 2} \text{PO}_4^{3-} + \text{4 } \mathrm{H_2O} \text{ (excess Ca(OH)₂)}\]
03

Write the Net Ionic Equation for First Neutralization

To write the net ionic equation, eliminate spectator ions from the balanced molecular equation:\[\text{2 } \mathrm{H_3PO_4} + \text{3 } \mathrm{Ca^{2+} + 6 } \mathrm{OH^-} \rightarrow \text{6 } \mathrm{H_2O} + \text{Ca_3(PO_4)_2} \text{ (s)} \]
04

Write the Net Ionic Equation for Second Neutralization

Further simplify the ionic form by eliminating spectating ions:\[\text{H}_2\text{PO}_4^- + \text{Ca^{2+} + 2 OH^-} \rightarrow \text{CaHPO}_4\text{ (aq) + H}_2\text{O} \]

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Key Concepts

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

Phosphoric Acid
Phosphoric acid, with the chemical formula H₃PO₄, is an inorganic acid commonly found in soft drinks and used in fertilizers. It contains three acidic hydrogen ions, making it a triprotic acid. This means it can donate one hydrogen ion at a time in a stepwise manner, resulting in successive neutralizations.
  • First, it loses a hydrogen ion to form dihydrogen phosphate ( ext{H}_2 ext{PO}_4^-).
  • Further loss leads to the formation of hydrogen phosphate ( ext{HPO}_4^{2-}).
  • Complete deprotonation results in phosphate ( ext{PO}_4^{3-}).
During acid-base reactions with bases, phosphoric acid's ability to donate multiple hydrogen ions is crucial. Each donation is a step in the neutralization process, enabling complete reaction depending on the amount and strength of the base present.
Calcium Hydroxide
Calcium hydroxide, Ca(OH)₂, is a strong base often referred to as slaked lime. It is used in construction (for mortar) and also to neutralize acidic soils in agriculture.

In neutralization reactions, calcium hydroxide works by releasing hydroxide ions (OH⁻), which then react with hydrogen ions (H⁺) from acids to form water (H₂O). This makes it an effective agent in neutralizing acidic compounds, like phosphoric acid.
  • Each molecule of Ca(OH)₂ provides two OH⁻ ions.
  • It can effectively neutralize multiple protons (H⁺) depending on the acid.
Calcium hydroxide also plays a role in ensuring complete neutralization when there are multiple equivalents of acidic hydrogen available, as seen with tripotic acids like H₃PO₄.
Net Ionic Equations
Net ionic equations are simplified versions of chemical equations that only show the particles that are directly involved in the chemical reaction. They eliminate spectator ions, which do not participate in the actual reaction. This simplification is critical in focusing on the substances that undergo change.

Here's a quick way to write them:
  • Write down the balanced molecular equation.
  • Identify and remove the spectator ions (those that appear unchanged on both sides).
  • What's left is the net ionic equation showing the real chemistry happening.
For the reaction of phosphoric acid and calcium hydroxide, we turn complex molecular equations into understandable net ionic forms. This process helps in visualizing the exchange of ions and developing an understanding of how substances interact at the ionic level.
Molecular Equations
Unlike net ionic equations, molecular equations represent all reactants and products as if they were intact molecules in their original states. This provides a comprehensive overview of all substances participating in a reaction without simplifying into ions and parts. They are essential for giving a full picture of the reactants leading to products.

In chemical notation:
  • The molecular equation outlines each compound as it starts and ends, preserved in molecular form.
  • Ensures understanding of the stoichiometry and the complete reaction process.
Using molecular equations for reactions between phosphoric acid and calcium hydroxide shows all stages of neutralization. They are especially useful for visualizing the full quantity and components involved before detailing them down to ions as seen in net ionic equations.

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Most popular questions from this chapter

An alloy of iron and carbon was treated with sulfuric acid, in which only iron reacts. $$ 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Fe}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{H}_{2}(g) $$ If a sample of alloy weighing \(2.358 \mathrm{~g}\) gave \(0.1067 \mathrm{~g}\) of hydrogen, what is the percentage of iron in the alloy?

Obtain the oxidation number for the element noted in each of the following. (a) \(\mathrm{Cr}\) in \(\mathrm{CrO}_{3}\) (b) \(\mathrm{Hg}\) in \(\mathrm{Hg}_{2} \mathrm{Cl}_{2}\) (c) Ga in \(\mathrm{Ga}(\mathrm{OH})_{3}\) (d) \(\mathrm{P}\) in \(\mathrm{Na}_{3} \mathrm{PO}_{4}\)

You are asked to prepare \(0.250 \mathrm{~L}\) of a solution that is \(0.500 M\) in nitrate ion. Your only source of nitrate ion is a bottle of \(1.00 M\) calcium nitrate. What volume (mL) of the calcium nitrate solution must you use?

A 25.0-mL sample of sodium sulfate solution was analyzed by adding an excess of barium chloride solution to produce barium sulfate crystals, which were filtered from the solution. \(\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{BaCl}_{2}(a q) \longrightarrow 2 \mathrm{NaCl}(a q)+\mathrm{BaSO}_{4}(s)\) If \(5.719 \mathrm{~g}\) of barium sulfate was obtained, what was the molarity of the original \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) solution?

Mercury(II) nitrate is treated with hydrogen sulfide, \(\mathrm{H}_{2} \mathrm{~S},\) forming a precipitate and a solution. Write the molecular equation and the net ionic equation for the reaction. An acid is formed; is it strong or weak? Name each of the products. If \(81.15 \mathrm{~g}\) of mercury(II) nitrate and \(8.52 \mathrm{~g}\) of hydrogen sulfide are mixed in \(550.0 \mathrm{~g}\) of water to form \(58.16 \mathrm{~g}\) of precipitate, what is the mass of the solution after the reaction?

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