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What is the pH of \(0.75 M \mathrm{H}_{2} \mathrm{SO}_{4} ?\)

Short Answer

Expert verified
Answer: The pH of a 0.75 M sulfuric acid solution is approximately -0.18.

Step by step solution

01

Calculate the concentration of H+ ions

As H2SO4 is a strong acid, it completely dissociates in water. The dissociation of H2SO4 in water is given by the equation: H2SO4 -> 2H+ + SO42- Since 1 mole of sulfuric acid dissociates into 2 moles of H+ ions, the concentration of H+ ions produced in the given 0.75 M H2SO4 solution can be calculated as follows: Concentration of H+ ions = 2 * (Concentration of H2SO4) Concentration of H+ ions = 2 * 0.75 M Concentration of H+ ions = 1.5 M
02

Calculate pH using the pH formula

Now that we have the concentration of H+ ions in the solution, we can calculate the pH using the formula: pH = -log10([H+]) Where [H+] is the concentration of H+ ions in moles per liter (M). Plugging in the calculated concentration of H+ ions: pH = -log10(1.5) pH 鈮 -0.18 The pH of a 0.75 M H2SO4 solution is approximately -0.18.

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Key Concepts

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

Strong Acids
In the world of chemistry, acids are substances that donate protons or hydrogen ions when dissolved in water. Among these, **strong acids** are particularly potent. A strong acid is defined by its ability to completely dissociate or break apart in a solution. This means that when you dissolve a strong acid in water, nearly 100% of the acid molecules release their hydrogen ions. This complete dissociation is what makes strong acids so reactive and impactful.

Common examples of strong acids include:
  • Hydrochloric acid (HCl)
  • Nitric acid (HNO鈧)
  • Sulfuric acid (H鈧係O鈧)
It is important in pH calculations because the concentration of hydrogen ions equals the concentration of the acid. For instance, if you have a 1M solution of a strong acid, the concentration of hydrogen ions produced will also be 1M.
Dissociation
Dissociation is a fundamental concept in chemistry. It involves the separation of molecules into smaller particles such as ions when placed in solution. For acids, dissociation is critically important because it defines how the acid behaves in a solution.

In the case of strong acids like sulfuric acid, they fully dissociate to produce hydrogen ions and other components. Using sulfuric acid as an example, when it dissociates in water, each molecule splits into two hydrogen ions and one sulfate ion (SO鈧劼测伝).

This process can be represented by the chemical equation:
  • H鈧係O鈧 鈫 2H鈦 + SO鈧劼测伝
Understanding dissociation is crucial for determining hydrogen ion concentration and subsequently calculating the pH of a solution. The more hydrogen ions produced, the more acidic the solution becomes, and thus, the lower the pH.
Sulfuric Acid
Sulfuric acid, with the chemical formula H鈧係O鈧, is one of the strongest and most widely used acids in industry and laboratories. Due to its high affinity for dissolving materials and other acids, it is a key player in numerous chemical processes.

In solution, sulfuric acid is known for its ability to dissociate completely. Unlike some other acids which lose their potency fully after the initial dissociation, sulfuric acid dissociates in two stages:
  • The first hydrogen ion is released easily.
  • The second hydrogen ion requires more energy to be removed.
This dual dissociation makes sulfuric acid particularly interesting compared to monoprotic strong acids. The ability to release two hydrogen ions per molecule is significant in reactions and when calculating pH as it effectively doubles the concentration of hydrogen ions in a solution.
Hydrogen Ion Concentration
To accurately determine the pH of a solution, one must understand the importance of hydrogen ion concentration, symbolized as [H鈦篯. It directly correlates to the level of acidity or basicity of the solution.

The concentration of hydrogen ions in an aqueous solution is crucial because pH is calculated based on this value. The formula used is:
  • pH = -log鈧佲個([H鈦篯)
The greater the concentration of hydrogen ions, the more acidic the solution is, translating to a lower pH value. Conversely, a lower concentration of hydrogen ions means the solution is more basic, resulting in a higher pH.

This concept was clearly demonstrated in the given exercise. Sulfuric acid's complete dissociation doubled the concentration of hydrogen ions to 1.5 M, leading to an extremely low pH value of -0.18. It's a key example of how strong acids can significantly alter hydrogen ion concentration and thus, the pH of a solution.

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Most popular questions from this chapter

Rhubarb Pie The leaves of the rhubarb plant contain high concentrations of diprotic oxalic acid (HOOCCOOH) and must be removed before the stems are used to make rhubarb pie. What is the pH of \(0.0288 M\) oxalic acid?

Identify the conjugate acid of each of the following species: \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{N}, \mathrm{CH}_{3} \mathrm{COO}^{-}, \mathrm{HSO}_{4}^{-},\) and \(\mathrm{OH}^{-}\)

Calculate the \(\mathrm{pH}\) and \(\mathrm{pOH}\) of the solutions with the following hydronium ion or hydroxide ion concentrations. Indicate which solutions are acidic, basic, or neutral. a. \(\left[\mathrm{OH}^{-}\right]=8.2 \times 10^{-11} \mathrm{M}\) b. \(\left[\mathrm{OH}^{-}\right]=7.7 \times 10^{-6} \mathrm{M}\) c. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=3.2 \times 10^{-4} \mathrm{M}\) d. \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=1.0 \times 10^{-7} \mathrm{M}\)

Describe a solution (solute and concentration) that has a negative pH value.

Predict which solution in each pair below will have the lower pH. a. \(2.56 \times 10^{-2} \mathrm{M} \mathrm{HCl}\) or \(4.09 \times 10^{-2} \mathrm{MHBr}\) b. \(1.00 \times 10^{-5} M\) acctic acid \(\left(K_{\mathrm{a}}=1.76 \times 10^{-5}\right)\) or \(1.00 \times 10^{-5} M\) formic acid \(\left(K_{a}=1.77 \times 10^{-4}\right)\) c. \(22 \mathrm{mM} \mathrm{CH}_{3} \mathrm{NH}_{2}\left(\mathrm{p} K_{\mathrm{b}}=3.36\right)\) or \(22 \mathrm{mM}\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\left(K_{\mathrm{b}}=5.9 \times 10^{-4}\right)\) d. \(158 \mathrm{mM} \mathrm{NH}_{3}\left(\mathrm{p} K_{\mathrm{b}}=4.75\right)\) or \(158 \mathrm{m} M\) acetic acid \(\left(p K_{a}=4.75\right)\) e. \(0.00395 M \mathrm{HNO}_{3}\) or \(0.00145 M \mathrm{HClO}_{4}\) f. \(2.05 \times 10^{-1} M\) propionic acid \(\left(K_{2}=1.4 \times 10^{-5}\right)\) or \(2.05 \times 10^{-1} M\) fluoroacetic acid \(\left(K_{2}=2.6 \times 10^{-3}\right)\) g. 375 m \(M\) pyridine \(\left(p K_{b}=8.77\right)\) or \(375 \mathrm{mM}\) aniline \(\left(\mathrm{p} K_{\mathrm{b}}=9.40\right)\) h. \(0.555 M \mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\left(K_{2}=3 \times 10^{-3}\right)\) or \(0.355 M \mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\left(K_{2}=1 \times 10^{-4}\right)\)

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