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Chapter 14: Problem 13

Consider the equation: $$ \mathrm{HA}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \leftrightharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{A}^{-}(a q) $$ a. If water is a better base than \(\mathrm{A}^{-},\) which way will the equilibrium lie? b. If water is a better base than \(\mathrm{A}^{-}\) , is HA a strong or a weak acid? c. If water is a better base than \(\mathrm{A}^{-},\) is the value of \(K_{\mathrm{a}}\) greater or less than 1 ?

Short Answer

Expert verified
a. The equilibrium will lie to the right since water is a better base and has a higher tendency to accept protons compared to A-. b. HA is a strong acid, as water's stronger basicity and the right-sided equilibrium favor its ionization. c. The value of Ka is greater than 1, as a higher Ka value correlates to stronger acid behavior and we have determined that HA is a strong acid.

Step by step solution

01

Determine the better base

In this case, we are given that water is a better base than A-. What this means is that water has a higher tendency to accept protons compared to A-. Therefore, in order to reach equilibrium, the reaction will favor the formation of H3O+ (aq) and A- (aq) since water successfully outcompetes A- for protons. Hence, the equilibrium lies to the right. **b. Nature of the acid HA**
02

Strong or Weak Acid

Since the equilibrium favors the right side of the reaction, indicating that HA is donating protons to water, we can conclude that HA is indeed an acid. However, we still need to determine whether it is a strong or weak acid. Due to water's stronger basicity compared to A- and the fact the equilibrium lies to the right, HA must be a strong acid. Strong acids tend to ionize readily in water, and in the given case, the nature of water allows it to do so. **c. Ka value of the reaction**
03

Comparing Ka value to 1

Ka stands for the acid dissociation constant, which measures the extent of dissociation of an acid. A larger Ka value indicates stronger acid behavior, while a smaller Ka value indicates weaker acid behavior. In our case, since we determined that HA is a strong acid, we can infer that its Ka value must be greater than 1. This is because strong acids have a greater tendency to ionize in water, in comparison to weak acids which have a Ka value lesser than 1.

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Most popular questions from this chapter

Are solutions of the following salts acidic, basic, or neutral? For those that are not neutral, write balanced equations for the reactions causing the solution to be acidic or basic. The relevant \(K_{\mathrm{a}}\) and \(K_{\mathrm{b}}\) values are found in Tables 14.2 and \(14.3 .\) $\begin{array}{ll}{\text { a. } \operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}} & {\text { d. } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{ClO}_{2}} \\ {\text { b. } \mathrm{NH}_{4} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}} & {\text { e. } \mathrm{NH}_{4} \mathrm{F}} \\ {\text { c. } \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{O} \mathrm{l}} & {\text { f. } \mathrm{CH}_{3} \mathrm{NH}_{3} \mathrm{CN}}\end{array}$

Acrylic acid \(\left(\mathrm{CH}_{2}=\mathrm{CHCO}_{2} \mathrm{H}\right)\) is a precursor for many important plastics. \(K_{\mathrm{a}}\) for acrylic acid is \(5.6 \times 10^{-5} .\) a. Calculate the pH of a \(0.10-M\) solution of acrylic acid. b. Calculate the percent dissociation of a \(0.10-M\) solution of acrylic acid. c. calculate the pH of a \(0.050-M\) solution of sodium acrylate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

For the reaction of hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) in water, $$ \mathrm{H}_{2} \mathrm{NNH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{NNH}_{3}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \(K_{\mathrm{b}}\) is \(3.0 \times 10^{-6} .\) Calculate the concentrations of all species and the pH of a \(2.0-M\) solution of hydrazine in water.

Would you expect \(\mathrm{Fe}^{3+}\) or \(\mathrm{Fe}^{2+}\) to be the stronger Lewis acid? Explain.

Classify each of the following as a strong acid, weak acid, strong base, or weak base in aqueous solution. a. \(\mathrm{HNO}_{2}\) b. HNO \(_{3}\) c. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) d. \(\mathrm{NaOH}\) e. \(\mathrm{NH}_{3}\) f. \(\mathrm{HF}\) g. \(\mathrm{HC}-\mathrm{OH}\) h. \(\mathrm{Ca}(\mathrm{OH})_{2}\) i. \(\mathrm{H}_{2} \mathrm{SO}_{4}\)
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