(Solved): Example 16.1 Identify the conjugate acid-base pairs in the reaction between ammonia and hydrofluoric ...

${\mathrm{NH}}_3(aq)+{\mathrm{HF}}^{(aq})={\mathrm{NH}}_4^{+}(aq)+{\mathrm{F}}^{?}(aq)$ Srrategy Remember that a conjugate base always has one fewer $\mathrm{H}$ atom and one more negative charge (or one fewer positive charge) than the formula of the corresponding acid. Solution ${\mathrm{NH}}_3$ has one fewer $\mathrm{H}$ atom and one fewer positive charge than ${\mathrm{NH}}_4^{+},{\mathrm{F}}^{?}$has one fewer $\mathrm{H}$ atom and one more negative charge than $\mathrm{HF}$. Therefore, the conjugate acid-base pairs are (1) ${\mathrm{NH}}_4^{+}$and ${\mathrm{NH}}_3$ and (2) ${\mathrm{HF}}_{\text{and }}$ F - Practice Exercise Identify the conjugate acid-base pairs for the reaction ${\mathrm{CN}}^{?}+{\mathrm{H}}_2\mathrm{O}?\mathrm{HCN}+{\mathrm{OH}}^{?}$ The concentration of ${\mathrm{OH}}^{?}$- ions in a certain household ammonia cleaning solution is $0.0030\mathrm{M}$. Calculate the concentration of ${\mathrm{H}}^{+}$ions. Strategy We are given the concentration of ${\mathrm{OH}}^{?}$ions and are asked to calculate $\left[{\mathrm{H}}^{+}\right]$. The relationship between $\left[{\mathrm{H}}^{+}\right]$and $\left[{\mathrm{OH}}^{?}\right]$in water or an aqueous solution is given by the ion-product constant of water, $K_{\mathrm{w}}[$ ] Equation $(16.4)]$ Solution Rearranging Squation (16.4), we write $?{\mathrm{H}}^{+}?=\frac{K_{\mathrm{W}}}{?{\mathrm{OH}}^{?}?}=\frac{1.0\times 10^{?14}}{0.0030}=3.3\times 10^{?12}\mathrm{M}$ Check Because $\left[{\mathrm{H}}^{+}\right]<\left[{\mathrm{OH}}^{?}\right]$, the solution is basic, as we would expect from the earlier discussion of the reaction of ammonis with water. Pracrice Exercise Calculate the concentration of ${\mathrm{OH}}^{?}$ions in a $\mathrm{HCl}$ solution whose hydrogen ion concentration is $1.3\mathrm{M}$. The concentration of ${\mathrm{H}}^{+}$ions in a bottle of table wine was $4.1\times 10^{?4}\mathrm{M}$ right after the cork was removed. Only half of the wine was consumed. The other half, after it had been standing open to the air for a month, was found to have a hydrogen ion concentration equal to $2.3\times 10^{?3}\mathrm{M}$. Calculate the $\mathrm{pH}$ of the wine on these two occasions. Srratecy. We are given the ${\mathrm{H}}^{+}$ion concentration and are asked to calculate the $\mathrm{pH}$ of the solution. What is the definition of $\mathrm{pH}$ ? Solution According to [?uation (16.5). $\mathrm{pH}=?\log \left[{\mathrm{H}}^{+}\right]$. When the bottle was first opened, $\left[{\mathrm{H}}^{+}\right]=4.1\times 10^{?4}M$, which we substitute in $?$ Equation (16.5): $\begin{array}{rl}\mathrm{pH}& =?\log ?{\mathrm{H}}^{+}?\\ & =?\log \left(4.1\times 10^{?4}\right)=3.39\end{array}$ On the second occasion, $\left[{\mathrm{H}}^{+}\right]=2.3\times 10^{?3}M$, so that $\mathrm{pH}=?\log \left(2.3\times 10^{?3}\right)=2.64$ In esch case, the pH has only two significant figures. The two digits to the right of the decimal in 3.39 teil us that there are fwo significant figures in the original number (see $(3$ Appendix 3 ) Comment The increase in hydrogen ion concentration (or decrease in $\mathrm{pH}$ ) is largely the result of the conversion of some of the alcohol (ethanol) to acetic acid, a reaction that takes place in the presence of molecular oxygen. Practice Exercise Nitric acid (HNO ) ${}_3$ is used in the production of fertilizer, dyes, drugs, and explosives, Caleulate the $\mathrm{pH}$ of a ${\mathrm{HNO}}_3$ solution having a hydrogen ion concentration of $\mathrm{0.76.}M$. The $\mathrm{pH}$ of rainwater collected in a certain region of the northeastern United States on a particular day was 5.23. Calculate the ${\mathrm{H}}^{+}$ion concentration of the rainwater. Strategy. Here we are given the $\mathrm{pH}$ of a solution and asked to calculate $\left[{\mathrm{H}}^{+}\right]$. Because $\mathrm{pH}$ is defined as $\mathrm{pH}=?\log \left[{\mathrm{H}}^{+}\right]$, we can solve for $\left[{\mathrm{H}}^{+}\right]$by taking the antilog of the $\mathrm{pH}$; that is, $\left[{\mathrm{H}}^{+}\right]=10^{?\mathrm{pH}}$, as shown in [?) Equation $(16,6)$. Solution From Equation (16.5). $\mathrm{pH}=?\log \left[{\mathrm{H}}^{+}\right]=5.23$ Therefore, $\log \left[{\mathrm{H}}^{+}\right]=?5.23$ To calculate $\left[{\mathrm{H}}^{+}\right]$, we need to take the antilog of -5.23 : $\left[{\mathrm{H}}^{+}\right]=10^{?5.21}=5.9\times 10^{?6}\mathrm{M}$ Scientific calcilators hove an antilog function that is sometimes labeled iNV log or $10^2$. Check Because the $\mathrm{pH}$ is between 5 and 6 , we can expect $\left[{\mathrm{H}}^{+}\right]$to be between $1\times 10^{?5}$ and $1\times 10^{?6}\mathrm{M}$. Therefore. the answer is reasonable. Practice Exercise The $\mathrm{pH}$ of a certain orange juice is 3.33. Calculate the ${\mathrm{H}}^{+}$ion concentration. In a $\mathrm{NaOH}$ solution $\left[{\mathrm{OH}}^{?}\right]$is $7.2\times 10^{?5}\mathrm{M}$. Calculate the $\mathrm{pH}$ of the solution. Srrategy Solving this problem takes two steps. First, we need to calculate the pOH using Equation (16.8). Next, we use ? Equation (16.10) to calculate the $\mathrm{pH}$ of the solution. Solution We use [?? Equation (16.8): $\begin{array}{rl}\mathrm{pOH}& =?\log \left[{\mathrm{OH}}^{?}\right]\\ & =?\log \left(7.2\times 10^{?5}\right)\\ & =4.14\end{array}$ Now we use [? Equation (16.10) $\begin{array}{rl}\mathrm{pH}+\mathrm{pOH}& =14.00\\ \mathrm{pH}& =14.00?\mathrm{pOH}\\ & =14.00?4.14=9.86\end{array}$ Alternatively, we can use the ion-product constant of water, $K_{\mathrm{w}}=\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{OH}}^{?}\right]$to calculate $\left[{\mathrm{H}}^{+}\right]$, and then we can calculate the $\mathrm{pH}$ from the $\left[{\mathrm{H}}^{+}\right]$. Try it. Check The answer shows that the solution is basic $(\mathrm{pH}>7)$, which is consistent with a $\mathrm{NaOH}$ solution. Pructice Exereise The ${\mathrm{OH}}^{?}$ion concentration of a blood sample is $2.5\times 10^{?7}\mathrm{M}$. What is the $\mathrm{pH}$ of the blood? Predict the relative strengths of the oxoacids in each of the following groups: (a) $\mathrm{HClO},\mathrm{HBrO}$, and $\mathrm{HIO}$ : (b) ${\mathrm{HNO}}_3$ and ${\mathrm{HNO}}_2$ Strategy Examine the molecular structure. In (a) the three acids bave similar structure but differ only in the central atom ( $\mathrm{Cl},\mathrm{Br}$, and I). Which central atom is the most electronegative? In (b) the acids have the same central atom ( $\mathrm{N}$ ) but differ in the number of $\mathrm{O}$ atoms. What is the oxidation number of $\mathrm{N}$ in each of these two acids? Solution (a) These acids all have the same structure, and the halogens all have the same oxidation number $(+1)$. Because the electronegativity decreases from $\mathrm{Cl}$ to $\mathrm{I}$, the polarity of the $\mathrm{X}?\mathrm{O}$ bond (where $\mathrm{X}$ denotes a halogen atom) increases from $\mathrm{HClO}$ to $\mathrm{HIO}$, and the polarity of the $\mathrm{O}?\mathrm{H}$ bond decreases from $\mathrm{HClO}$ to $\mathrm{HIO}$. Thus, the acid strength decreases as follows: $\mathrm{HClO}>\mathrm{HBrO}>\mathrm{HIO}$ (b) The structures of ${\mathrm{HNO}}_3$ and ${\mathrm{HNO}}_2$ are shown in $\mathrm{G}$ Figure 16.5. Because the oxidation number of $\mathrm{N}$ is +5 in ${\mathrm{HNO}}_3$ and +3 in ${\mathrm{HNO}}_2,{\mathrm{HNO}}_3$ is a stronger acid than ${\mathrm{HNO}}_2$. Practice Exercise Which of the following acids is weaker: ${\mathrm{HBrO}}_3$ or ${\mathrm{HBrO}}_4$ ? Predict whether the following solutions will be acidic, basic, or nearly neutral: (a) ${\mathrm{NH}}_4\mathrm{I}$, (b) ${\mathrm{NaNO}}_2,(\mathrm{c}){\mathrm{FeCl}}_3$, (d) ${\mathrm{NH}}_4\mathrm{F}$. Strategy In deciding whether a salt will undergo hydrolysis, ask yourself the following questions: Is the cation a highly charged metal ion or an ammonium ion? Is the anion the conjugate base of a weak acid? If yes to either question. then hydrolysis will occur. In cases where both the cation and the anion react with water, the $\mathrm{pH}$ of the solution will depend on the relative magnitudes of $K_{\mathrm{a}}$ for the cation and $K_{\mathrm{b}}$ for the anion (see [? Table 16.7). Solution We first break up the salt into its cation and anion components and then examine the possible reaction of each ion with water. (a) The cation is ${\mathrm{NH}}_4^{+}$, which will hydrolyze to produce ${\mathrm{NH}}_3$ and ${\mathrm{H}}^{+}$. The ${\mathrm{I}}^{?}$anion is the conjugate base of the strong acid HI. Therefore. ${\mathrm{I}}^{?}$will not hydrolyze and the solution is acidic. (b) The ${\mathrm{Na}}^{+}$cation does not hydrolyze. The ${\mathrm{NO}}_2^{?}$is the conjugate base of the weak acid ${\mathrm{HNO}}_2$ and will hydrolyze to give ${\mathrm{HNO}}_2$ and ${\mathrm{OH}}^{?}$. The solution will be basic. (c) ${\mathrm{Fe}}^{3+}$ is a small metal ion with a high charge and hydrolyzes to produce ${\mathrm{H}}^{+}$ions. The ${\mathrm{Cl}}^{?}$does not hydrolyze. Consequently, the solution will be acidic. (d) Both the ${\mathrm{NH}}_4^{+}$and ${\mathrm{F}}^{?}$ions will hydrolyze. From $\mathrm{C}$ Tables 16.5 and 16.3 we see that the $K_3$ of ${\mathrm{NH}}_4^{+}\left(5.6\times 10^{?10}\right)$ is greater than the $K_b$ for ${\mathrm{F}}^{?}\left(1.4\times 10^{?11}\right)$. Therefore, the solution will be acidic. Practice Exercise Predict whether the following solutions will be acidic, basic, or nearly neutral: (a) LiClO 4 , (b) Na ${\mathrm{PO}}_4$, (c) $\mathrm{Bi}{\left({\mathrm{NO}}_3\right)}_3$, (d) ${\mathrm{NH}}_4\mathrm{CN}$. 16.14 Define $\mathrm{pOH}$. Write an equation relating $\mathrm{pH}$ and $\mathrm{pOH}$. 16.15 Calculate the hydrogen ion concentration for solutions with these $\mathrm{pH}$ values: (a) 2.42, (b) 11.21, (c) 6.96, (d) 15.00 . 16.16 Calculate the hydrogen ion concentration in moles per liter for each of these solutions: (a) a solution whose pH is 5.20 , (b) a solution whose $\mathrm{pH}$ is 16.00 , (c) a solution whose hydroxide concentration is $3.7\times 10^{?9}\mathrm{M}$.