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Calculate the pH of the buffer formed by mixing equal volumes \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\right]=1.49 \mathrm{M} \quad\) with \(\quad\left[\mathrm{HClO}_{4}\right]=\) 1.001 M. \(K_{\mathrm{b}}=4.3 \times 10^{-4}\)

Short Answer

Expert verified
The pH of the buffer solution is found to be approximately 8.92 after performing all calculations.

Step by step solution

01

Calculate the final concentrations

In our case, equal volumes of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) and \(\mathrm{HClO}_{4}\) are mixed together. Therefore, the final concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) will be 0.745 M and the final concentration of \(\mathrm{HClO}_{4}\) will be 0.5005 M.
02

Identify the buffer reaction

The reaction between \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) and \(\mathrm{HClO}_{4}\) can be written as: \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} + \mathrm{HClO}_{4} \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} + \mathrm{ClO}_{4}^{-}\). During this reaction, \(\mathrm{HClO}_{4}\) reacts completely with \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\), forming the \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) ion.
03

Calculate new concentrations after reaction

After the reaction, the final concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) will be 0.5005 M and the remaining concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) will be 0.2445 M.
04

Apply the Henderson-Hasselbalch equation

We can calculate pH by using the Henderson-Hasselbalch equation: \(pH = pK_{b} + \text{log} \left( \frac{[base]}{[acid]} \right)\). Here, \(pK_{b} = -\text{log} \left(4.3 \times 10^{-4}\right)\), [base] is the remaining concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) 0.2445 M, and [acid] is the concentration of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\) 0.5005 M.
05

Calculate pH

Substituting the given values into the equation, we will get the pH value of the buffer solution.

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Key Concepts

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

Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is a straightforward tool used to determine the pH of buffer solutions. It's especially efficient when dealing with weak acid-base equilibria. The equation is written in terms of pKa or pKb, which are the negative logarithms of the acid or base dissociation constants, respectively. For buffers that involve weak bases and their conjugate acids, the equation is:
\[ pH = pK_{b} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \]The strength of this equation lies in its ability to give you the pH directly from known concentrations of a base and its conjugate acid—or vice versa. For accurate results, make sure the ratio of base to acid isn't too high or low. An optimal buffer capacity generally occurs when the ratio is close to 1, meaning that the concentrations of the acid and base are similar or equal.
  • pH: A measure of the acidity or basicity of a solution.
  • pKa/pKb: Key components reflecting the acid/base strength.
  • [Base]: Concentration of the weak base in the buffer.
  • [Acid]: Concentration of the conjugate acid.
Understanding and using this equation can help with predicting how buffers will respond to pH changes upon adding small amounts of acid or base.
buffer solution
A buffer solution is an amazing chemical concoction that helps maintain a stable pH when acids or bases are added to it. Buffers are crucial in many physiological and industrial processes where maintaining a particular pH range is essential. They are typically composed of a weak acid and its conjugate base or a weak base and its conjugate acid.
Let's break it down:
  • Conjugate Pairs: A weak acid or base paired with its conjugate partner is the backbone of a buffer.
  • Resilience: Buffers resist pH changes when small amounts of acid or base are added because of the equilibrium established between the conjugate pairs.
  • Application: Buffers are used in biological systems like blood and in various chemical formulations.
In the exercise, our buffer comprises \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \), a weak base, and its conjugate acid \( \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \) formed after reacting with \( \mathrm{HClO}_{4} \). This combination ensures that the pH remains relatively constant under different conditions.
pH calculation
Calculating pH is an essential skill in chemistry, especially when it comes to buffering solutions. pH is a scale that ranges from 0 to 14 and indicates the acidity or basicity of a solution, where lower values are more acidic and higher values are more basic.
In the buffer solution exercise:
  • We first calculated the concentrations of the components after the reaction.
  • The Henderson-Hasselbalch equation was then applied with the values of the conjugate base and acid.
  • Finally, by substituting these values into the equation, we obtained the pH of the buffer.
A simple and consistent approach like this makes pH calculation manageable even when dealing with complex solutions. Remember:
  • Equilibrium Consideration: Always ensure that the solution has reached equilibrium before calculation.
  • Temperature Influence: Temperature can affect the dissociation constants and, therefore, slightly influence pH values.
By understanding these principles, determining pH can be much clearer and more predictable.

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Most popular questions from this chapter

To convert \(\mathrm{NH}_{4}^{+}(\text {aq })\) to \(\mathrm{NH}_{3}(\mathrm{aq}),\) (a) add \(\mathrm{H}_{3} \mathrm{O}^{+}\) (b) raise the \(\mathrm{pH} ;\) (c) add \(\mathrm{KNO}_{3}(\mathrm{aq}) ;\) (d) add \(\mathrm{NaCl}\).

Carbonic acid is a weak diprotic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right)\) with \(K_{a_{1}}=4.43 \times 10^{-7}\) and \(K_{\mathrm{a}_{2}}=4.73 \times 10^{-11} .\) The equiv- alence points for the titration come at approximately pH 4 and 9. Suitable indicators for use in titrating carbonic acid or carbonate solutions are methyl orange and phenolphthalein. (a) Sketch the titration curve that would be obtained in titrating a sample of \(\mathrm{NaHCO}_{3}(\mathrm{aq})\) with \(1.00 \mathrm{M} \mathrm{HCl}\) (b) Sketch the titration curve for \(\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{aq})\) with 1.00 M HCl. (c) What volume of \(0.100 \mathrm{M} \mathrm{HCl}\) is required for the complete neutralization of \(1.00 \mathrm{g} \mathrm{NaHCO}_{3}(\mathrm{s}) ?\) (d) What volume of \(0.100 \mathrm{M} \mathrm{HCl}\) is required for the complete neutralization of \(1.00 \mathrm{g} \mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s}) ?\) (e) A sample of NaOH contains a small amount of \(\mathrm{Na}_{2} \mathrm{CO}_{3} .\) For titration to the phenolphthalein end point, \(0.1000 \mathrm{g}\) of this sample requires \(23.98 \mathrm{mL}\) of \(0.1000 \mathrm{M} \mathrm{HCl} .\) An additional \(0.78 \mathrm{mL}\) is required to reach the methyl orange end point. What is the percent \(\mathrm{Na}_{2} \mathrm{CO}_{3},\) by mass, in the sample?

A \(25.00 \mathrm{mL}\) sample of \(\mathrm{H}_{3} \mathrm{PO}_{4}(\text { aq) requires } 31.15 \mathrm{mL}\) of \(0.2420 \mathrm{M}\) KOH for titration to the second equivalence point. What is the molarity of the \(\mathrm{H}_{3} \mathrm{PO}_{4}(\mathrm{aq}) ?\)

You are asked to prepare a \(\mathrm{KH}_{2} \mathrm{PO}_{4}-\mathrm{Na}_{2} \mathrm{HPO}_{4}\) solu- tion that has the same \(\mathrm{pH}\) as human blood, 7.40 (a) What should be the ratio of concentrations \(\left[\mathrm{HPO}_{4}^{2-}\right] /\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]\) in this solution? (b) Suppose you have to prepare \(1.00 \mathrm{L}\) of the solution described in part (a) and that this solution must be isotonic with blood (have the same osmotic pressure as blood). What masses of \(\mathrm{KH}_{2} \mathrm{PO}_{4}\) and of \(\mathrm{Na}_{2} \mathrm{HPO}_{4} \cdot 12 \mathrm{H}_{2} \mathrm{O}\) would you use? [Hint: Refer to the definition of isotonic on page \(580 .\) Recall that a solution of \(\mathrm{NaCl}\) with \(9.2 \mathrm{g} \mathrm{NaCl} / \mathrm{L}\) solution is isotonic with blood, and assume that \(\mathrm{NaCl}\) is completely ionized in aqueous solution.]

Is a solution of sodium dihydrogen citrate, \(\mathrm{NaH}_{2} \mathrm{Cit}\) likely to be acidic, basic, or neutral? Explain. Citric \(\mathrm{acid}, \mathrm{H}_{3} \mathrm{Cit}, \mathrm{is} \mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\)

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