Chapter 8: Acids and base (C9961647)

 #toc { border: 1px solid #bba; background-color: #f7f8ff; padding: 1em; font-size: 90%; text-align: center; } #toc-header { display: inline; padding: 0; font-size: 100%; font-weight: bold; } #toc ul { list-style-type: none; margin-left: 0; padding-left: 0; text-align: left; } .toc2 { margin-left: 1em; } .toc3 { margin-left: 2em; } .toc4 { margin-left: 3em; }Table of Contents Last modified: 1265d agoWord count: 2,986 wordsLegend: Key principles // Storyline

Getting started

“We see all those signs about how dangerous acids are,” Mandy started, “why are they dangerous?”

“They are highly corrosive,” Jamie replied.

$\ce{H+}$ is regarded as the shorthand of the hydronium $\ce{H3O^+}$ ion, a hydrated proton, because bare protons do not exist as free species in aqueous solution. Whereas acid tastes sour [like lemon juice], bases taste bitter [like soup]. Bases are slippery, and is how soaps work.

Acid strength is measured by a pH scale, where $\ce{pH= -log[H+]}$. Analogously, $\ce{pOH= -log[OH-]}$, and $\ce{pK_{a}=-log[K_{a}]}$. A pH of 7 at $25^{\circ}C$ is neutral, meaning $\ce{[H+]=[OH-]}$. A pH of <7 at $25^{\circ}C$ is acidic, meaning $\ce{[H+]>[OH-]}$. A pH of >7 at $25^{\circ}C$ is basic, meaning $\ce{[H+]>[OH-]}$. Note that the concentration of $\ce{H+}$ ions, and the pH, seems to be inverse, which is the effect of the negative log function. The pH is by in large determined by the exponent of the $\ce{[H+]}$, for example, $10^{-12}$ is a pH of 12, $10^{-2}$ is a pH of 2.

If an acid $HA$ is added to water, the associated chemical equation is $\ce{HA_{(aq)}+H2O_{(l)} -> H3O+_{(aq)} + A^-_{(aq)}}$. The conjugate base of $HA$ is $A^-$, because it is the acid ($HA$) without a proton [so it can now accept]. The conjugate acid of $H2O$ is $H3O+$, because it is the base ($H2O$) with an extra proton [which it can now donate]. If a base $A-$ is added to water, the associated chemical equation is $\ce{A-_{(aq)}+H2O_{(l)} -> HA_{(aq)} + OH-_{(aq)} }$. Here, the conjugate base of $\ce{H2O}$ is $\ce{OH^{-}}$, and the conjugate acid of $\ce{A-}$ is $HA$. Note therefore, that water has both a conjugate acid and base, and is known as amphoteric, as it can act as both an acid and base.

The factors determining acid strength include:

• Bond polarity, which if high, creates a strong acid. Because an acid is defined by the Lewis definition as an electron pair acceptor, increased bond polarity creates a greater electronegativity difference between the binding elements. Therefore, for example, if bound to the slightly positive hydrogen, a highly electronegative element will have greater ability to accept electron pairs. This explains why the oxyacids are more acidic with increasing oxygens, because even though their bond strength (discussed ) is high, the bond polarity dominates
• Bond strength, which if weak, creates a strong acid. Because an acid is defined by the Bronsted-Lowry definition as the ability to donate protons, if its bond strength is weaker, it can donate these protons more easily. This is because if bond strength is less, the size of the atoms sharing the bond is also less. For the acid $Ha$, as the size of the atom $A$ increases, the strength of the bond decreases, thereby increasing acidity. This explains why hydrofluoric acid $HF$ is considered a weak acid, even though it has high bond polarity (just ). Because the atomic radius of fluorine is much smaller, it shares a stronger bond with hydrogen, than other hydrohalic acids ($HCl, HBr, HI$), which in fact increase in acidity going down the periodic table, because the size of the atoms become significantly larger, dominating its acidity
• Stability of the conjugate base, which if stable, creates a strong acid

Polyprotic acids are acids that have more than one proton, and can therefore donate more than one proton. The second [and so forth] protons are so weak, its effect on pH is negligible, the exception being sulfuric acid $\ce{H2SO4}$, especially in dilute solutions. The second proton dissociates more easily in dilute solutions than concentrated solutions, because even though the amount of molecules dissociating in concentrated solutions is greater [than dilute solutions], the percent dissociation in dilute solutions is more.

1 Acid/base equilibrium

Self-ionization of water is the reaction of water with itself, $\ce{2H2O \rightleftharpoons H3O+ + OH-}$. Because water is far more stable than the ions produced, this is evidently an endothermic reaction (heat is required, absorbed, discussed ). Therefore, the equilibrium constant [found through the law of mass action, $K=\dfrac{[S]^\sigma.[T]^\tau}{[A]^\alpha.[B]^\beta}$, which in this circumstance is $\ce{K=[H}+.[OH]-}$], describing the ratio of concentrations [of products over reactants] when an equilibrium is reached, is very small, emphasizing that there is far less products than reactants, in fact, $K_{w}=10^{-14}$ at $25^{\circ}C$. Since $pK_{w}=-log[K_{w}]$, $pK_{w}=14$. Note that $pH+pOH=pK_{w}=14$, based on the mathematical rules of logarithms.

If acid is added to water $\ce{HA+H2O \rightleftharpoons H+ + A-}$, the equilibrium constant will be $\ce{K_{a}=\dfrac{[H+][A-]}{[HA]}}$, known as the acid dissociation constant. Note that a larger $K_{a}$, and therefore a lower $pK_{a}$, indicates a stronger acid. Note that water is not present [as ], as it is a pure liquid. If base is added to water $\ce{A- + H2O \rightleftharpoons HA + OH-}$, the equilibrium constant will be $\ce{K_{b}=\dfrac{[HA][OH-]}{[A-]}}$. Note that $\ce{K_{a}.K_{b}=\dfrac{[H+][A-]}{[HA]}.\dfrac{[HA][OH-]}{[A-]}=[H]+.[OH]-=pH+pOH=pK_{w}=14}$.

Strong acids [and bases] completely dissociate in water. This means that the $\ce{H+}$ concentration of the acidic solution, is equal to the $\ce{H+}$ concentration of the acid. For example, a 0.01M solution of HCl, has a pH of $-log[0.01]=2$. Another example, a 0.01M solution of NaOH, has a pOH of $-log[0.01]=2$. Since $pH+pOH=14$, therefore, $pH=14-2=12$. Examples of strong acids include the hydrogen halides (except hydrofluoric acid, so therefore include $HCl, HI, HBr$), and $\ce{HClO4, HClO3, HNO3, H2SO4}$. Examples of strong bases include the hydroxides of the [Group 1] alkali metals ($NaOH, KOH$), and the hydroxides of the [Group 2] alkaline earth metals ($\ce{Ca(OH)2}$). Magnesium hydroxide $\ce{Mg(OH)2}$ has low solubility in water (discussed ).

To find the pH of weak acids undergoing dissociation ($\ce{HA+H2O\rightleftharpoons H+ + A-}$), the use of $K_{a}$ is necessary, $\ce{K_{a}=\dfrac{[H+][A-]}{[HA]}}$, where $\ce{[H+]}$ is the concentration of $\ce{H+}$ ions, $\ce{[A-]}$ is the concentration of conjugate base ions, and $[HA]$ is the concentration of non-dissociated acid molecules. At equilibrium, we know that:

Therefore, the acid dissociation constant can be rewritten as $K_{a}=\dfrac{x.x}{C_{HA}-x}$, or reshuffling and simplifying, $x^2+K_{a}x-K_{a}C_{HA}=0$. Since $\ce{[H+]=x}$, this formula can be rewritten as $\ce{[H+]^2+K_{a}[H+]-K_{a}C_{HA}=0}$. This can be solved using the quadratic formula, $x=-b \pm\dfrac{\sqrt{b^2-4ac}}{2a}$, where $\ce{x=[H+], a=1, b=K_{a}}$, and $c=-K_{a}C_{HA}$, meaning the formula becomes $\ce{[H+]=-K_{a} \pm\dfrac{\sqrt{2K_{a}+4K_{a}C_{HA}} }{2} }$. If the degree of dissociation is quite small, note that $x[latex] [or [latex]\ce{[H+]}$] $<< C_{HA}$, meaning that the $C_{HA}$ will dominant. This reduces the formula to $\ce{[H+]\approx \dfrac{\sqrt{4K_{a}C_{HA} } }{2} }$. For example, the pH of 0.01M of benzoic acid (which has $pK_{a}=4.19$), $C_{HA}=0.01$ and $K_{a}=4.19$, $\ce{[H+]\approx \sqrt{0.01\times 10^{-4.19} }=8\times 10^{-4} }$. Therefore, $\ce{pH=-log{[H+]}\approx 3.1}$.

To find the pH of weak bases, $K_{b}$ is used to find $pOH$, which can then be converted to $pH$.

2 Titration

Titration is a technique used to determine unknown concentration, by dripping a strong acid or base, known as the titrant, into the unknown solution. Titration curve is the plot of pH of the solution [on the y-axis], against the volume of the titrant added [on the x-axis], and is of a sigmoid S-shape. There is a beginning horizontal region, a vertical region, and then another horizontal region. The middle of the steep vertical region, known as an inflection point, is the equivalence [or stoichiometric] point. This is where the number of moles of acid/base in the titrant added, is equal to the number of moles of acid/base originally in the solution. © 2014 MR. SHUM

The half-equivalence point is the middle of the flat horizontal region, and indicates the point halfway between the beginning of the curve and the equivalence point. At this point, the number of moles of acid/base added by the titrant, is equal to [as it dissociates] half the number of moles of acid/base originally in the solution. [Whereas at the equivalence point, where $HA$ has entirely dissociated into $\ce{A-}$], at the half way half-equivalence point, only half the $HA$ has dissociated into $\ce{A-}$, meaning that the number of moles of $HA$ equals the number of moles of $\ce{A-}$, as they are equal. Note that during this flat horizontal region, when acid or base is added, the pH doesn’t change much, and is known as a buffer solution. The pH of a buffer solution can be found with the Henderson-Hasselbalch equation, which is $\ce{pH=pK_{a}+log{\dfrac{A-}{HA} }}$. Given that at the half-equivalence point, $\ce{[HA]=[A-]}$, we can in-substitute, and reduce, to find that $pH=pK_{a}$. At the equivalence point, remember that it was just stated that $HA$ has almost entirely dissociated into $\ce{A-}$, meaning that $[HA]\approx 0$. This would result in an infinitely large pH. Rather, we use the Heylamn equation, which is expressed in terms of the $pK_{b}$ of $\ce{A-}$. Remember the relationship $K_{a}.K_{b}=K_{w}$, or reshuffling, $K_{b}=\dfrac{K_{w}}{K_{a}}$. Therefore, rather than finding the pH of $[HA]$, we find the pOH of $\ce{[A-]}$.

“Oh, I love the Hoff!” Mandy commented.

“Not Hasselhoff: Hasselbalch,” Jamie replied, “Having a Sophie moment there ?”

“JAMIE!!!!” Mandy replied, “Well she is blonde , and you know Jamie, brunettes have more fun.”

Indicators change colors based on pH, and can therefore confirm arrival at the equivalence point, as there is a large change in pH at this steep vertical region of the titration curve, and therefore a large change in color. Indicators are weak acids or bases, and so their colors tend to only show a particular pH range, that may be $\pm1-2 pH$ from its $pKa$. With highly irregular curves, pH meters may be required, which convert a voltage into a pH reading, and can therefore read pH at any and every point.

Titration curve for polyprotic acids are analogous to those for monoprotic acids, except instead of just having one half-equivalence and equivalence point, there is one for each proton. This demonstrates that all of the first acid’s first most acidic proton is removed, before the second most acidic proton, and so forth.

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