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A \(0.100 M\) solution of bromoacetic acid \(\left(\mathrm{BrCH}_{2} \mathrm{COOH}\right)\) is \(13.2 \%\) ionized. Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{BrCH}_{2} \mathrm{COO}^{-}\right],\left[\mathrm{BrCH}_{2} \mathrm{COOH}\right]\) and \(K_{a}\) for bromoacetic acid.

Short Answer

Expert verified
The concentrations of the species at equilibrium are: \([\mathrm{H}^{+}]\) = 0.0132 M, \([\mathrm{BrCH}_{2}\mathrm{COO}^{-}]\) = 0.0132 M, and \([\mathrm{BrCH}_{2} \mathrm{COOH}]\) = 0.0868 M. The ionization constant, \(K_a\), for bromoacetic acid is 0.00200.

Step by step solution

01

Write the ionization equation

The ionization of bromoacetic acid can be represented as follows: BrCH2COOH (aq) ⇌ H+ (aq) + BrCH2COO- (aq)
02

Determine initial concentrations

We are given that the initial concentration of BrCH2COOH is 0.100 M, so our starting concentrations are: [BrCH2COOH] = 0.100 M [H+] = 0 M [BrCH2COO–] = 0 M
03

Determine the change in concentrations due to ionization

Since the equilibrium is 13.2% ionized, we know that the concentration of H+ and BrCH2COO- at equilibrium is 13.2% of the initial concentration of BrCH2COOH (0.100 M). Therefore, the change in concentration for H+ and BrCH2COO- is: Δ[H+] = Δ[BrCH2COO–] = 0.100 M × 0.132 = 0.0132 M
04

Calculate the equilibrium concentrations

Now we need to find the equilibrium concentrations by calculating: [BrCH2COOH]_eq = initial [BrCH2COOH] - Δ[BrCH2COOH] = 0.100 M - 0.0132 M = 0.0868 M [H+]_eq = initial [H+] + Δ[H+] = 0 M + 0.0132 M = 0.0132 M [BrCH2COO–]_eq = initial [BrCH2COO–] + Δ[BrCH2COO–] = 0 M + 0.0132 M = 0.0132 M
05

Calculate Ka using the equilibrium concentrations

Now we'll use the equilibrium constant expression: Ka = \(\frac{[\mathrm{H}^{+}][\mathrm{BrCH}_{2}\mathrm{COO}^{-}]}{[\mathrm{BrCH}_{2}\mathrm{COOH}]}\) Plug in the equilibrium concentrations: Ka = \(\frac{(0.0132)(0.0132)}{(0.0868)}\) = 0.00200 The calculated value for Ka is 0.00200. To summarize, we have calculated: [H+] = 0.0132 M [BrCH2COO–] = 0.0132 M [BrCH2COOH] = 0.0868 M Ka = 0.00200

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Key Concepts

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

Equilibrium Concentrations
In chemistry, when dealing with weak acids like bromoacetic acid, we often discuss something called "equilibrium concentrations."
This term refers to the concentrations of all reactants and products in a reaction after it has reached balance, or "equilibrium." At equilibrium, there is no further change in the concentration of each species.

For bromoacetic acid, when dissolved in water, it partially ionizes into hydrogen ions (H extsuperscript{+}) and bromide ions (BrCH extsubscript{2}COO extsuperscript{-}).
  • The solution initially starts with a certain concentration of bromoacetic acid, here it is 0.100 M.
  • Since the acid ionizes by 13.2%, the change in concentration of the ions can be calculated using this percentage.
This is how we determine the subsequent concentrations at equilibrium:

- The concentration of bromoacetic acid decreases.
- The concentrations of H extsuperscript{+} and BrCH extsubscript{2}COO extsuperscript{-} increase by 13.2% of the initial acid concentration.

Once these changes are accounted for, we compute the precise equilibrium concentrations, ensuring we understand the dynamics of the chemical balance better.
Equilibrium Constant (Ka)
The equilibrium constant, denoted as Ka, is a particular kind of constant used to express the extent of ionization of a weak acid.This measure gives us insight into how much an acid dissociates into its ions in a solution.
A larger Ka value means the acid ionizes more completely, indicating a stronger acid, whereas a smaller Ka indicates weaker ionization.

Formally, the equilibrium constant for the acid dissociation reaction is determined by the equation:
\[ K_a = \frac{[H^+][BrCH_2COO^-]}{[BrCH_2COOH]}\]

Here, the products of ionization, H extsuperscript{+} and BrCH extsubscript{2}COO extsuperscript{-}, appear in the numerator, and the undissociated acid, BrCH extsubscript{2}COOH, is in the denominator. The calculated Ka for bromoacetic acid, using the equilibrium concentrations we determined, is 0.00200, reflecting its partial ionization behavior.
Bromoacetic Acid
Bromoacetic acid, with the chemical formula BrCH extsubscript{2}COOH, is an organic compound and a member of the haloacetic acids family. It is distinguished by the presence of a bromine atom attached to the acetic acid skeleton, which influences its chemical properties significantly.

This acid is classified as a weak acid because it does not fully dissociate in water. Instead, it establishes an equilibrium between ionized and non-ionized forms.
  • This characteristic is typical for many acids with Ka values smaller than about 1.
  • Such partial ionization can be quantified precisely, as done in calculating equilibrium concentrations and the equilibrium constant, Ka.

Understanding these properties of bromoacetic acid not only helps in quantitative analysis but is also essential for applications where precise pH control is needed.

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Most popular questions from this chapter

(a) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as a base in \(\mathrm{H}_{2} \mathrm{O}(l)\) (b) Write an equation for the reaction in which \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q)\) acts as an acid in \(\mathrm{H}_{2} \mathrm{O}(l) .(\mathrm{c})\) What is the conjugate acid of \(\mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{7} \mathrm{O}_{5}^{-}(a q) ?\) What is its conjugate base?

Calculate the percent ionization of hydrazoic acid \(\left(\mathrm{HN}_{3}\right)\) in solutions of each of the following concentrations \(\left(K_{a}\right.\) is given (b) \(0.100 M,(c) 0.0400 M\). in Appendix \(D\) ): (a) \(0.400 M\),

(a) Given that \(K_{b}\) for ammonia is \(1.8 \times 10^{-5}\) and that for hydroxylamine is \(1.1 \times 10^{-8}\), which is the stronger base? (b) Which is the stronger acid, the ammonium ion or the hydroxylammonium ion? (c) Calculate \(K_{a}\) values for \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{H}_{3} \mathrm{NOH}^{+}\)

Identify the Lewis acid and Lewis base in each of the following reactions: (a) \(\mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{FeBr}_{3}(s)+\mathrm{Br}^{-}(a q) \rightleftharpoons \mathrm{FeBr}_{4}^{-}(a q)\) (c) \(\mathrm{Zn}^{2+}(a q)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)\) (d) \(\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\)

Indicate whether each of the following statements is correct or incorrect. For those that are incorrect, explain why they are wrong. (a) Every Bronsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Bronsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) \(\mathrm{K}^{+}\) ion is acidic in water because it causes hydrating water molecules to become more acidic. (e) The percent ionization of a weak acid in water increases as the concentration of acid decreases.

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