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What is the \(\mathrm{pH}\) of a \(0.1500 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) solution if (a) the ionization of \(\mathrm{HSO}_{4}^{-}\) is ignored? (b) the ionization of \(\mathrm{HSO}_{4}^{-}\) is taken into account? \(\left(K_{\mathrm{a}}\right.\) for \(\mathrm{HSO}_{4}^{-}\) is \(\left.1.1 \times 10^{-2} .\right)\)

Short Answer

Expert verified
Question: Calculate the pH of a 0.1500 M H2SO4 solution considering two scenarios: (a) ignoring the ionization of HSO4- and (b) taking into account the ionization of HSO4-. Given the Ka value for HSO4- is 1.1 x 10^-2. Answer: a) pH (ignoring HSO4鈭 ionization) = 0.52 b) pH (accounting for HSO4鈭 ionization) = 0.50

Step by step solution

01

Write the dissociation equation of H2SO4

H2SO4 fully dissociates in water to produce 2 H+ ions and an SO4^2- ion. $$ \mathrm{H}_{2}\mathrm{SO}_{4} \rightarrow 2\mathrm{H}^{+} + \mathrm{SO}_{4}^{2-} $$ Since we are ignoring the ionization of HSO4-, all H+ ions are generated from the first ionization of H2SO4.
02

Calculate the concentration of H+ ions

Since H2SO4 dissociates into 2 H+ ions for each molecule, the concentration of H+ ions is twice the concentration of H2SO4. $$ [\mathrm{H}^{+}] = 2 \times [\mathrm{H}_{2}\mathrm{SO}_{4}] $$ Substitute the given concentration of H2SO4 (0.1500 M) into the equation. $$ [\mathrm{H}^{+}] = 2 \times 0.1500 $$ Calculate the concentration of H+ ions. $$ [\mathrm{H}^{+}] = 0.3000 \mathrm{M} $$
03

Calculate the pH

Now use the relationship between pH and the concentration of H+ ions to find the pH. $$ \mathrm{pH} = -\log [\mathrm{H}^{+}] $$ Substitute the concentration of H+ ions found in Step 2. $$ \mathrm{pH} = -\log 0.3000 $$ Calculate the pH. $$ \textbf {pH (ignoring HSO4鈭 ionization)} = 0.52 $$ #b) Scenario 2: Taking into account the ionization of HSO4-#
04

Write the dissociation equation of HSO4-

Now, we need to consider the ionization of HSO4- in the solution. $$ \mathrm{HSO}_{4}^{-} \rightleftharpoons \mathrm{H}^{+} + \mathrm{SO}_{4}^{2-} $$
05

Write the expression for Ka

Write the expression for the Ka of HSO4- and substitute the given Ka value. $$ K_{\mathrm{a}} = \frac{[\mathrm{H}^{+}][\mathrm{SO}_{4}^{2-}]}{[\mathrm{HSO}_{4}^{-}]} = 1.1 \times 10^{-2} $$
06

Set up an ICE table

Set up an ICE (Initial, Change, Equilibrium) table to analyze the ionization of HSO4-. The change in the concentration of H+ will be represented as x, and we are starting at the H+ concentration found in Scenario 1, Step 2 (0.3000 M) due to the dissociation of H2SO4. ``` | HSO4- | H+ | SO4^2- ---------------------------- Initial | 0.1500 | 0.3 | 0 Change | -x | +x | +x ---------------------------- Equilibrium | 0.1500-x | 0.3000+x | x ```
07

Substitute the equilibrium concentrations into the Ka expression and solve for x

Substitute the equilibrium concentrations from the ICE table into the Ka expression and solve for x. $$ 1.1 \times 10^{-2} = \frac{(0.3000 + x)(x)}{0.1500 - x} $$ Solve the equation for x to get the additional concentration of H+ ions generated from the ionization of HSO4-. Due to the quadratic nature of the equation, it is easiest to solve using a calculator or spreadsheet. \(x = 0.0167 \mathrm{M}\)
08

Calculate the final concentration of H+ ions and the pH

Add the additional concentration of H+ ions from the ionization of HSO4- to the initial concentration of H+ ions from the dissociation of H2SO4. $$ [\mathrm{H}^{+}]_{total} = 0.3000 + 0.0167 = 0.3167 \mathrm{M} $$ Now use the relationship between pH and the concentration of H+ ions to find the pH. $$ \mathrm{pH} = -\log [\mathrm{H}^{+}_{total}] $$ Substitute the total concentration of H+ ions. $$ \mathrm{pH} = -\log 0.3167 $$ Calculate the pH. $$ \textbf {pH (accounting for HSO4鈭 ionization)} = 0.50 $$

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Key Concepts

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

Acid Dissociation
When we talk about acid dissociation, we are referring to the process where an acid breaks apart in water to form ions, typically releasing hydrogen ions (H鈦) into the solution. For strong acids like sulfuric acid ( H鈧係O鈧 ), the dissociation is almost complete, meaning that nearly all the acid molecules ionize to release hydrogen ions. This step is crucial in understanding how acids increase the concentration of hydrogen ions in a solution, which directly affects the ph level. Understanding acid dissociation helps in calculating the pH of a solution precisely, especially when dealing with strong acids that fully dissociate and weak acid forms that only partially ionize. This distinction is important when dealing with acids like HSO鈧勨伝, which acts as a conjugate base with a smaller Ka value, indicating its partial ionization.
Sulfuric Acid
Sulfuric acid ( H鈧係O鈧 ) is a strong and common mineral acid with significant industrial and laboratory uses. It is known for its ability to fully dissociate in water, releasing two protons (H鈦) per molecule. This property makes sulfuric acid a diprotic acid, capable of undergoing two dissociation steps in solution:
  • First Dissociation: H鈧係O鈧 鈫 2H鈦 + SO鈧刕{2-}
  • Second Dissociation: HSO鈧勨伝 鈬 H鈦 + SO鈧刕{2-}
In the first dissociation, H鈧係O鈧 is a strong acid, ionizing completely. However, during the second dissociation, HSO鈧勨伝 acts as a weaker acid and only partially dissociates, depending on the equilibrium conditions. This dual behavior of H鈧係O鈧 affects the total concentration of hydrogen ions in the solution, which further influences the pH calculations.
Equilibrium Constant
The equilibrium constant (Ka) is a measure of the tendency of a compound, often an acid, to dissociate in solution. For the ionization of HSO鈧勨伝, the equilibrium constant signifies how readily it releases hydrogen ions. The Ka value is derived from the equilibrium concentrations of the ions in the solution and is a crucial factor when dealing with weak acids, including the second dissociation of sulfuric acid.The expression for Ka of a reaction is written as:\[Ka = \frac{[Products]}{[Reactants]}\]For HSO鈧勨伝:\[Ka = \frac{[H鈦篯[SO鈧刕{2-}]}{[HSO鈧勨伝]}\]Here, knowing the Ka value helps predict the position of equilibrium and thus determine the degree to which HSO鈧勨伝 contributes to the total hydrogen ion concentration. By solving the equilibrium equation, one can calculate the additional H鈦 ion concentration due to HSO鈧勨伝 ionization, which is necessary to accurately determine the pH.
Ionization
Ionization refers to the process where an atom or a molecule acquires a negative or positive charge by gaining or losing electrons. In the context of acids, it typically involves the release of hydrogen ions (H鈦) into a solution. This concept is central to understanding why acids can alter the pH of a solution. Strong acids, like sulfuric acid ( H鈧係O鈧 ), fully ionize in water, making them particularly powerful at increasing H鈦 concentration. For a given acid like H鈧係O鈧 :
  • Primary Ionization: Involves the complete dissociation of the first hydrogen, which happens readily.
  • Secondary Ionization: Involves the partial ionization of HSO鈧勨伝 . This step is influenced by the equilibrium constant and defines the weak acid behavior part of a diprotic acid.
The ionization process is crucial for determining how much of the acid contributes to the H鈦 ion pool, impacting the overall pH of the solution significantly. By understanding ionization, chemists can predict and calculate changes in pH when dealing with various acids in solution.

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Most popular questions from this chapter

A buffer is prepared using the propionic acid/propionate \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2} /\right.\) \(\left.\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}\right)\) acid-base pair for which the ratio \(\left[\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right] /\left[\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-}\right]\) is \(4.50 .\) \(K_{\mathrm{a}}\) for propionic acid is \(1.4 \times 10^{-5}\). (a) What is the \(\mathrm{pH}\) of this buffer? (b) Enough strong base is added to convert \(27 \%\) of \(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) to \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}^{-} .\) What is the \(\mathrm{pH}\) of the resulting solution? (c) Strong base is added to increase the \(\mathrm{pH}\). What must the acid/base ratio be so that the \(\mathrm{pH}\) increases by exactly one unit (e.g., from 2 to 3 ) from the answer in (a)?

A \(20.00\) -mL sample of \(0.220 \mathrm{M}\) triethylamine, (CH \(\left._{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}\), is titrated with \(0.544 \mathrm{M} \mathrm{HCl}\). $$\left(\mathrm{Kb}\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}=5.2 \times 10^{-4}\right)$$ (a) Write a balanced net ionic equation for the titration. (b) How many \(\mathrm{mL}\) of \(\mathrm{HCl}\) are required to reach the equivalence point? (c) Calculate \(\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{~N}\right],\left[\left(\mathrm{CH}_{3} \mathrm{CH}_{2}\right)_{3} \mathrm{NH}^{+}\right],\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{Cl}^{-}\right]\) at the equivalence point. (Assume that volumes are additive.) (d) What is the \(\mathrm{pH}\) at the equivalence point?

Explain why (a) the pH decreases when lactic acid is added to a sodium lactate solution. (b) the \(\mathrm{pH}\) of \(0.1 \mathrm{M} \mathrm{NH}_{3}\) is less than \(13.0\). (c) a buffer resists changes in pH caused by the addition of \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-} .\) (d) a solution with a low \(\mathrm{pH}\) is not necessarily a strong acid solution.

There is a buffer system \(\left(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}-\mathrm{HPO}_{4}{ }^{2-}\right)\) in blood that helps keep the blood \(\mathrm{pH}\) at about \(7.40 .\left(\mathrm{K}_{\mathrm{a}} \mathrm{H}_{2} \mathrm{PO}_{4}^{-}=6.2 \times 10^{-8}\right)\). (a) Calculate the \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right] /\left[\mathrm{HPO}_{4}^{2-}\right]\) ratio at the normal \(\mathrm{pH}\) of blood. (b) What percentage of the \(\mathrm{HPO}_{4}{ }^{2-}\) ions are converted to \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) when the \(\mathrm{pH}\) goes down to \(6.80\) ? (c) What percentage of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ions are converted to \(\mathrm{HPO}_{4}{ }^{2-}\) when the \(\mathrm{pH}\) goes up to \(7.80\) ?

A buffer is prepared by dissolving \(0.037\) mol of potassium fluoride in \(135 \mathrm{~mL}\) of \(0.0237 \mathrm{M}\) hydrofluoric acid. Assume no volume change after KF is dissolved. Calculate the \(\mathrm{pH}\) of this buffer.

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