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Neutralization reactions involving either a strong acid or a strong base go essentially to completion, and therefore we must take such neutralizations into account before calculating concentrations in mixtures of acids and bases. Consider a mixture of \(3.28 \mathrm{~g}\) of \(\mathrm{Na}_{3} \mathrm{PO}_{4}\) and \(300.0 \mathrm{~mL}\) of \(0.180 \mathrm{M} \mathrm{HCl} .\) Write balanced net ionic equations for the neutralization reactions, and calculate the \(\mathrm{pH}\) of the solution.

Short Answer

Expert verified
The pH of the solution is approximately 6.22.

Step by step solution

01

Calculate Moles of \(\mathrm{Na}_3\mathrm{PO}_4\)

First, calculate the moles of \(\mathrm{Na}_3\mathrm{PO}_4\) using its molar mass. \(\mathrm{Na}_3\mathrm{PO}_4\) has a molar mass of 163.94 g/mol. Moles of \(\mathrm{Na}_3\mathrm{PO}_4\) = \( \frac{3.28 \, \mathrm{g}}{163.94 \, \mathrm{g/mol}} \approx 0.020 \, \mathrm{mol} \).
02

Calculate Moles of \(\mathrm{HCl}\)

Next, calculate the moles of \(\mathrm{HCl}\) in the solution. Volume = 300.0 mL = 0.300 L, and concentration = 0.180 M, hence moles of \(\mathrm{HCl}\) = \(0.180 \, M \times 0.300 \, L = 0.054 \, \mathrm{mol}\).
03

Write Balanced Net Ionic Equations

The reaction between \(\mathrm{Na}_3\mathrm{PO}_4\) and \(\mathrm{HCl}\) can be simplified into dissolution of \(\mathrm{Na}_3\mathrm{PO}_4\) into \(\mathrm{PO}_4^{3-}\) ions, and neutralization with \(\mathrm{H}^+\) ions from \(\mathrm{HCl}\):\(\mathrm{PO}_4^{3-} + 3\mathrm{H}^+ \rightarrow \mathrm{H}_3\mathrm{PO}_4}\)
04

Determine Limiting Reactant

Using the stoichiometry from the ionic equation, every mole of \(\mathrm{PO}_4^{3-}\) reacts with 3 moles of \(\mathrm{H}^+\). Moles of \(\mathrm{PO}_4^{3-}\) = 0.020 mol and needs \(0.060 \, \mathrm{mol}\) of \(\mathrm{H}^+\). Since only 0.054 mol \(\mathrm{H}^+\) is available, \(\mathrm{HCl}\) is the limiting reactant.
05

Calculate Remaining \(\mathrm{PO}_4^{3-}\)

Since \(\mathrm{HCl}\) is limiting, all \(\mathrm{H}^+\) ions will be used to neutralize \(\mathrm{PO}_4^{3-}\). Remaining \(\mathrm{PO}_4^{3-}: (0.020 \, \mathrm{mol} - \frac{0.054 \, \mathrm{mol}}{3}) = 0.002 \, \mathrm{mol}\).
06

Calculate \(\mathrm{OH}^{-}\) from \(\mathrm{PO}_4^{3-}\)

\(\mathrm{PO}_4^{3-}\) can act as a weak base. Use the equilibrium expression for \(\mathrm{PO}_4^{3-}\): \[ \mathrm{PO}_4^{3-} + \mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{HPO}_4^{2-} + \mathrm{OH}^-\]Assume \([\mathrm{OH}^-] = x\), and use \(K_b = \frac{K_w}{K_a}\) where \(K_w = 1.0 \times 10^{-14}\) and \(K_a \approx 7.1 \times 10^{-3}\) to find \(K_b\approx 1.4 \times 10^{-12}\).
07

Approximate \([\mathrm{OH}^-]\)

Assume \(K_b \approx x^2/ 0.002 \rightarrow x = \sqrt{K_b \times 0.002}\). Solve: \(x = \sqrt{1.4 \times 10^{-12} \times 0.002} \approx 1.67 \times 10^{-8} \mathrm{M}\).
08

Calculate \(\mathrm{pH}\)

Use \([\mathrm{OH}^-]\) to find \(\mathrm{pOH}\) and from it, \(\mathrm{pH}\): \(\mathrm{pOH} = -\log(1.67 \times 10^{-8}) \approx 7.78\),then \(\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 7.78 = 6.22\).

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Key Concepts

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

Strong Acids
Strong acids are substances that completely ionize in solution. In other words, they release all of their hydrogen ions, (H鈦), when dissolved in water. Hydrochloric acid (HCl) is an example of a strong acid commonly used in neutralization reactions. When HCl dissolves in water, it releases H鈦 ions, helping to lower the pH of the solution.

Strong acids have a very low pH value, nearing 0. When you consider neutralization reactions, strong acids play a critical role. They need to be completely neutralized by a base to prevent an unsafe level of acidity. This makes them react fully with bases, meaning they are incorporated into nearly every possible reaction type when mixed with a base.
  • Involves complete dissociation.
  • Low pH value.
  • Common in industrial and laboratory settings.
Strong Bases
Strong bases are the counterparts to strong acids; they completely dissociate in water to provide hydroxide ions, (OH鈦). An example of a strong base is sodium hydroxide (NaOH). It dissociates into sodium ions (Na鈦) and hydroxide ions, providing a high pH solution.

In neutralization reactions, strong bases effectively raise the pH by donating OH鈦 ions, which react with H鈦 ions to form water. This process neutralizes the solution. The neutralization capacity of a strong base is important, making them key reactants in many chemical reactions where controlling pH is critical.
  • Completely dissociates into OH鈦 ions.
  • Typically has a high pH, approaching 14.
  • Commonly used to neutralize acids in chemical processes.
Net Ionic Equations
Net ionic equations are simplified representations of reactions where only the particles that undergo a change are shown. Unreactive ions, sometimes called spectator ions, are left out. This simplification helps illustrate the essence of a chemical reaction, like neutralization.

In the neutralization reaction between Na鈧働O鈧 and HCl, the net ionic equation involves assembling the ions that actually change, like phosphate ions reacting with hydrogen ions to form H鈧働O鈧. Writing balanced net ionic equations helps students focus on key parts of the reaction without getting lost in unnecessary details.
  • Focuses on reacting species.
  • Ions that don't participate are excluded.
  • Clarifies the core of the chemical process.
pH Calculation
pH is a measure of the acidity or alkalinity of a solution, defined as the negative logarithm of the hydrogen ion concentration (\( \text{pH} = -\log[\text{H}^+] \) ). Calculating pH involves understanding the balance between H鈦 ions and OH鈦 ions. The neutral point on the pH scale is 7, where concentrations of H鈦 and OH鈦 are equal.

In the given reaction, you initially determine how many moles of OH鈦 are formed from the reaction. Using this concentration of OH鈦, you then find the pOH first, which represents the power of the hydroxide ion concentration ( \( \text{pOH} = -\log[\text{OH}^-] \) ). Finally, you convert pOH to pH using the relationship \( \text{pH} = 14 - \text{pOH} \). This step-by-step procedure gives insight into the chemical balance present in solutions.
  • Measures hydrogen ion concentration.
  • pH + pOH = 14.
  • Neutral solution = pH 7.

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Most popular questions from this chapter

We've said that alkali metal cations do not react appreciably with water to produce \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions, but in fact, all cations are acidic to some extent. The most acidic alkali metal cation is the smallest one, \(\mathrm{L}^{+}\), which has \(K_{\mathrm{a}}=2.5 \times 10^{-14}\) for the reaction \(\mathrm{Li}\left(\mathrm{H}_{2} \mathrm{O}\right)_{4}^{+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{Li}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}(\mathrm{OH})(a q)\) This reaction and the dissociation of water must be considered simultaneously in calculating the \(\mathrm{pH}\) of \(\mathrm{Li}^{+}\) solutions, which nevertheless have \(\mathrm{pH}=7 .\) Check this by calculating the \(\mathrm{pH}\) of \(0.10 \mathrm{M} \mathrm{LiCl} .\)

The concentration of \(\mathrm{OH}^{-}\) in human blood is \(2.24 \times 10^{-7} \mathrm{M} .\) Calculate the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) ions, and classify the solution as acidic, neutral, or basic.

Give the formula for the conjugate base of each of the following Bronsted- Lowry acids: (a) \(\mathrm{HSO}_{4}^{-}\) (b) \(\mathrm{H}_{2} \mathrm{SO}_{3}\) (c) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (d) \(\mathrm{NH}_{4}^{+}\) (e) \(\mathrm{H}_{2} \mathrm{O}\) (f) \(\mathrm{NH}_{3}\)

Write a balanced net ionic equation and the corresponding equilibrium equation for the reaction of the following weak bases with water: (a) Pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}\) (b) Ethylamine, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) (c) Acetate ion, \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}\)

Aqueous solutions of hydrogen sulfide contain \(\mathrm{H}_{2} \mathrm{~S}, \mathrm{HS}^{-}, \mathrm{S}^{2-}\). \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{OH}^{-}\), and \(\mathrm{H}_{2} \mathrm{O}\) in varying concentrations. Which of these species can act only as an acid? Which can act only as a base? Which can act both as an acid and as a base?

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