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Chapter 16: Problem 54

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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Most popular questions from this chapter

Codeine \(\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}\right)\) is a weak organic base. A \(5.0 \times 10^{-3} \mathrm{M}\) solution of codeine has a \(\mathrm{pH}\) of \(9.95 .\) Calculate the value of \(K_{b}\) for this substance. What is the \(\mathrm{p} K_{b}\) for this base?

Arrange the following \(0.10 \mathrm{M}\) solutions in order of increasing acidity (decreasing pH): (i) \(\mathrm{NH}_{4} \mathrm{NO}_{3}\) (ii) \(\mathrm{NaNO}_{3}\), (iii) $$ \mathrm{CH}_{3} \mathrm{COONH}_{4} \text { , (iv) } \mathrm{NaF} \text { , (v) } \mathrm{CH}_{3} \mathrm{COONa} \text { . } $$

Complete the following table by calculating the missing entries. In each case indicate whether the solution is acidic or basic. $$ \begin{array}{llll} \hline\left[\mathrm{H}^{+}\right] & \mathrm{OH}^{-} \text {(aq) } && \mathrm{pH} & &\text { pOH }&& \text { Acidic or basic? } \\ \hline 7.5 \times 10^{-3} \mathrm{M} & & & & \\ & 3.6 \times 10^{-10} \mathrm{M} & & \\ & && 8.25 & & \\ & & &&& 5.70 & \end{array} $$

Calculate the molar concentration of \(\mathrm{OH}^{-}\) ions in a \(0.075 \mathrm{M}\) solution of ethylamine \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} ; K_{b}=6.4 \times 10^{-4}\right) .\) Calcu- late the \(\mathrm{pH}\) of this solution.

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) \(\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{SO}_{3}(a q)\)
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