Acetic acid is a weak acid, and its dissociation reaction is written as:

\(\displaystyle HC_{2}H_{3}O_{2} + H_{2}O \rightleftharpoons C_{2}H_{3}O_{2}^{-} + H_{3}O^{+}\)

\(\displaystyle K_{a} = \frac{[C_{2}H_{3}O_{2}^-][H_{3}O^{+}]}{[HC_{2}H_{3}O_{2}]}\)

Using an ICE table, we can find the pH of the solution when the sodium acetate is added:

I: There is initially a 0.50M concentration of acetic acid. Because sodium acetate will dissolve completely, there will be a 0.050M concentration of acetate ions in the solution.

C: As acetic acid dissociates, the concentrations of acetate ions and hydronium ions will increase by an unknown concentration \(\displaystyle x\). Conversely, the acetic acid concentration will decrease by the same amount.

E: By setting the equilibrium expression equal to the acid equilibrium constant, we can solve for the value of \(\displaystyle x\).

\(\displaystyle K_{a} = \frac{[C_{2}H_{3}O_{2}^-][H_{3}O^{+}]}{[HC_{2}H_{3}O_{2}]}\)

\(\displaystyle 1.8*10^{-5} = \frac{(0.05+x)(x)}{(0.50-x)}\)

*Because the value for \(\displaystyle x\) will be so much lower than the initial concentration, we can omit it from the acetate expression as well as the acetic acid concentration:

\(\displaystyle 1.8*10^{-5} = \frac{(0.05)(x)}{(0.5)}\)

\(\displaystyle x=1.8*10^{-4}\)

Keep in mind that \(\displaystyle x\) is equal to the concentration of hydronium ions now in the solution.

\(\displaystyle [H_3O^+]=1.8*10^{-4}M\)

Use this value in the equation for pH:

\(\displaystyle pH=-log[H_3O^+]=-log(1.8*10^{-4})=3.74\)

The acid dissociation constant (Ka) is used to distinguish strong acids from weak acids. Strong acids have exceptionally high Ka values.

The Ka value is found by looking at the equilibrium constant for the dissociation of the acid. The higher the Ka, the more the acid dissociates. Thus, strong acids must dissociate more in water. In contrast, a weak acid is less likely to ionize and release a hydrogen ion, thus resulting in a less acidic solution.

\(\displaystyle HA\rightarrow H^++A^-\)

\(\displaystyle K_a=\frac{[H^+][A^-]}{[HA]}\)

This equation makes it clear that the more the acid converts from its original form to its ionzied, dissociated form, the higher the Ka value will be.

The Ka is the acid dissociation constant, and thus it is what determines how strong the acid is. Stronger acids dissociate to a greater extent and produce lower pH values.

The definition of \(\displaystyle K_a\) for \(\displaystyle HIO_3^-\) is:

\(\displaystyle K_a=0.16=\frac{[H^+][IO_3^-]}{[HIO_3]}\)

Set up an "ICE" table:

\(\displaystyle \begin{vmatrix} & H^+& IO_3^-& HIO_3\\ I&0& 0& .0045\\ C&X &X &-X \\ E& X &X &.0045-X \end{vmatrix}\)

Plug in values:

\(\displaystyle K_a=.16=\frac{[X][X]}{[.0045-X]}\)

\(\displaystyle X^2+.16X-.00072=0\)

Use the quadratic formula to solve:

\(\displaystyle X=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\)

\(\displaystyle X=.00438\)

\(\displaystyle [IO_3^-]=.00438M\)

\(\displaystyle K_a=1.76*10^{-5}=\frac{[H^+][CH_3COO^-]}{[CH_3COOH]}\)

Set up an "ICE" table

\(\displaystyle \begin{vmatrix} & H^+& CH_3COO^-& CH_3COOH\\ I&0& 0& .0027\\ C&X &X &-X \\ E& X &X &.0027-X \end{vmatrix}\)

Plug in values:

\(\displaystyle K_a=1.76*10^{-5}=\frac{[X][X]}{[.0027-X]}\)

\(\displaystyle X^2+1.76*10^{-5}X-4.75*10^{-8}=0\)

Use the quadratic formula:

\(\displaystyle X=2.09*10^{-4}\)

\(\displaystyle [CH_3COO^-]=2.09*10^{-4}M\)

In order to find the base dissociation constant for the conjugate base, we can start by finding the acid dissociation constant for the acid. Since a 1M solution of the acid has a pH of 4.6, we can find the proton concentration of the solution.

\(\displaystyle pH=-\log[H^+]\rightarrow [H^+]=10^{-pH}\)

\(\displaystyle \small [H^{+}] = 10^{-4.6} = 2.51*10^{-5}\)

Since the acid is monoprotic, we can set the following equilibrium expression equal to its acid dissociation constant.

\(\displaystyle \small K_{a} = \frac{[H^{+}][A^{-}]}{[HA]}\)

We can see that, since the acid is monoprotic, the concntration of protons will be equal to the concentration of the acid anion. The final concentration of the acid molecule will be equal to the initial concentration, minus the amount of protons formed. Using these values, we can solve for the equilibrium constant for the acid.

\(\displaystyle K_a= \frac{[2.51*10^{-5}]^{2}}{[1-2.51*10^{-5}]} = 6.31*10^{-10}\)

Now that we have the acid dissociation constant, we can find the conjugate base's dissociation constant by setting the product of the two values equal to the autoionization of water.

\(\displaystyle \small K_{w} = K_{a}K_{b}\)

\(\displaystyle \small 1*10^{-14} = K_{b}(6.31*10^{-10})\)

\(\displaystyle \small K_{b} = 1.58*10^{-5}\)

A weak acid/base best buffers about 1 pH point above and below its pKa. The pKA closest to the middle of 4 and 6 (so want as close to 5) is acetic acid at 4.7.

Nitric Acid is a strong acid and can't buffer. Rubidium Hydroxide is a strong base and thus can't buffer. Of the remaining, both are weak acids, but the one with a greater concentration has a greater buffering capacity.

The definition of a buffer solution is that it contains a weak acid and its conjugate base, or a weak base and its conjugate acid. Since we are starting with a weak acid in this case, we need its conjugate base.

To answer this question we need to look at the reaction below:

\(\displaystyle H_2CO_3_(_a_q_)+H_2O_(_l_)\rightleftharpoons HCO_3^-_(_a_q_) + H_3O^+_{(aq)}\)

An increase in the pH will result in a decrease in the concentration of hydrogen ions (\(\displaystyle H_3O^+\)). Using Le Chatelier’s principle we can find out which answer choices will decrease \(\displaystyle [H_3O^+]\).

Removing carbonic acid will decrease the concentration of \(\displaystyle H_2CO_3\). To maintain equilibrium, the reaction will shift to the left and make more reactants from products; therefore, there will be a decrease in the \(\displaystyle [H_3O^+]\) and an increase in pH.

Recall that salts like sodium bicarbonate, or \(\displaystyle NaHCO_3\), will dissociate in water and form ions. Sodium bicarbonate will form sodium (\(\displaystyle Na^+\)) and bicarbonate (\(\displaystyle HCO_3^-\)) ions. This side reaction will result in an increase in the bicarbonate ion concentration. Le Chatelier’s principle will shift the equilibrium of the given reaction to the left and, therefore, decrease the \(\displaystyle [H_3O^+]\). Adding sodium bicarbonate will increase the pH.