Chapter 14, Problem 47QRT

(a)

Interpretation Introduction

Interpretation:

Without doing any calculations, the $pH$ of $0.10M$ aqueous $HN{O}_2$ solution has to be assigned from the given $pH$ ranges: $pH2$, $pH$ between $2and6$, $pH$ between $6and8$, $pH$ between $8and12$, $pH12.$

Concept Introduction:

Strong acid with a concentration of $0.10M$ will have $pH1.$  The $pH$ range of weak acids will be between $2and6.$  Strong bases with a concentration of $0.10M$ will have $pH13.$ The $pH$ range of weak bases will be between $8and12.$

Explanation of Solution

$HN{O}_2$ is a weak acid as its $K_a$ value is $7.1\times {10}^{-4}.$  Its $pH$ will be between $2and6.$

(b)

Interpretation Introduction

Interpretation:

Without doing any calculations, the $pH$ of $0.10M$ aqueous $N{H}_{4}Cl$ solution has to be assigned from the given $pH$ ranges: $pH2$, $pH$ between $2and6$, $pH$ between $6and8$, $pH$ between $8and12$, $pH12.$

Concept Introduction:

Refer to part (a).

Explanation of Solution

$N{H}_{4}Cl$ is a salt consisting of $N{H}_4{}^{+}$ ion and $C{l}^{-}$ ion.  $N{H}_4{}^{+}$ ion is a weak acid and its $K_b$ value is $5.6\times {10}^{-10}.$  The $C{l}^{-}$ ions do not affect the $pH$ of a water solution.  Thus, its $pH$ will be between $2and6.$

  $N{H}_{4}Cl\left(s\right)\to N{H}_4{}^{+}\left(aq\right)+C{l}^{-}\left(aq\right)$

(c)

Interpretation Introduction

Interpretation:

Without doing any calculations, the $pH$ of $0.10M$ aqueous $NaF$ solution has to be assigned from the given $pH$ ranges: $pH2$, $pH$ between $2and6$, $pH$ between $6and8$, $pH$ between $8and12$, $pH12.$

Concept Introduction:

Refer to part (a).

Explanation of Solution

$NaF$ is a salt consisting of $N{a}^{+}$ ion and $F^{-}$ ion.  $F^{-}$ ion is a weak base and its $K_b$ value is $1.5\times {10}^{-11}.$  The $N{a}^{+}$ ions do not affect the $pH$ of a water solution.  Thus, its $pH$ will be between $8and12.$

  $NaF\left(s\right)\to N{a}^{+}\left(aq\right)+F^{-}\left(aq\right)$

(d)

Interpretation Introduction

Interpretation:

Without doing any calculations, the $pH$ of $0.10M$ aqueous $Mg{\left(C{H}_{3}COO\right)}_2$ solution has to be assigned from the given $pH$ ranges: $pH2$, $pH$ between $2and6$, $pH$ between $6and8$, $pH$ between $8and12$, $pH12.$

Concept Introduction:

Refer to part (a).

Explanation of Solution

$Mg{\left(C{H}_{3}COO\right)}_2$ is a salt consisting of $M{g}^{2+}$ ion and $C{H}_{3}CO{O}^{-}$ ion.  $C{H}_{3}CO{O}^{-}$ ion is a weak base and its $K_b$ value is $5.6\times {10}^{-10}.$  The $M{g}^{2+}$ ions do not affect the $pH$ of a water solution.  Thus, its $pH$ will be between $8and12.$

  $Mg{\left(C{H}_{3}COO\right)}_2\left(s\right)\to M{g}^{2+}\left(aq\right)+2C{H}_{3}CO{O}^{-}\left(aq\right)$

(e)

Interpretation Introduction

Interpretation:

Without doing any calculations, the $pH$ of $0.10M$ aqueous $BaO$ solution has to be assigned from the given $pH$ ranges: $pH2$, $pH$ between $2and6$, $pH$ between $6and8$, $pH$ between $8and12$, $pH12.$

Concept Introduction:

Refer to part (a).

Explanation of Solution

$BaO$ is a salt consisting of $B{a}^{2+}$ ion and $O^{2-}$ ion.  $O^{2-}$ ion is a strong base.  The $B{a}^{2+}$ ions do not affect the $pH$ of a water solution.  Thus, its $pH$ will be $12.$

  $BaO\left(s\right)\to B{a}^{2+}\left(aq\right)+O^{2-}\left(aq\right)$

(f)

Interpretation Introduction

Interpretation:

Without doing any calculations, the $pH$ of $0.10M$ aqueous $KHS{O}_4$ solution has to be assigned from the given $pH$ ranges: $pH2$, $pH$ between $2and6$, $pH$ between $6and8$, $pH$ between $8and12$, $pH12.$

Concept Introduction:

Refer to part (a).

Explanation of Solution

$KHS{O}_4$ is a salt consisting of $K^{+}$ ion and $HS{O}_4{}^{-}$ ion.  The $K^{+}$ ions do not affect the $pH$ of a water solution.  $HS{O}_4{}^{-}$ ion is a weak acid and a weak base.  Thus, the comparison of the value of $K_{a}and{K}_b$ should be done.

  $\begin{array}{l}KHS{O}_4\left(s\right)\to K^{+}\left(aq\right)+HS{O}_4{}^{-}\left(aq\right)\\ \\ HS{O}_4{}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+S{O}_4{}^{2-}\left(aq\right)K_a=1.1\times {10}^{-2}\\ \\ HS{O}_4{}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_{2}S{O}_4\left(aq\right)+O{H}^{-}\left(aq\right)K_b=verysmall\end{array}$

$HS{O}_4{}^{-}$ is a stronger acid than it is a base, thus its $pH$ will be between $2and6.$

(g)

Interpretation Introduction

Interpretation:

Without doing any calculations, the $pH$ of $0.10M$ aqueous $NaHC{O}_3$ solution has to be assigned from the given $pH$ ranges: $pH2$, $pH$ between $2and6$, $pH$ between $6and8$, $pH$ between $8and12$, $pH12.$

Concept Introduction:

Refer to part (a).

Explanation of Solution

$NaHC{O}_3$ is a salt consisting of $N{a}^{+}$ ion and $HC{O}_3{}^{-}$ ion.  The $N{a}^{+}$ ions do not affect the $pH$ of a water solution.  $HC{O}_3{}^{-}$ ion is a weak acid and a weak base.  Thus, the comparison of the value of $K_{a}and{K}_b$ should be done.

  $\begin{array}{l}NaHC{O}_3\left(s\right)\to N{a}^{+}\left(aq\right)+HC{O}_3{}^{-}\left(aq\right)\\ \\ HC{O}_3{}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+C{O}_3{}^{2-}\left(aq\right)K_a=4.7\times {10}^{-11}\\ \\ HC{O}_3{}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_{2}C{O}_3\left(aq\right)+O{H}^{-}\left(aq\right)K_b=2.3\times {10}^{-8}\end{array}$

$HC{O}_3{}^{-}$ is a stronger base than it is an acid, thus its $pH$ will be between $8and12.$

(h)

Interpretation Introduction

Interpretation:

Without doing any calculations, the $pH$ of $0.10M$ aqueous $BaC{l}_2$ solution has to be assigned from the given $pH$ ranges: $pH2$, $pH$ between $2and6$, $pH$ between $6and8$, $pH$ between $8and12$, $pH12.$

Concept Introduction:

Refer to part (a).

Explanation of Solution

$BaC{l}_2$ is a salt consisting of $B{a}^{2+}$ ion and $C{l}^{-}$ ion.  Neither of these  ions affect the $pH$ of a water solution.  Thus, its $pH$ will be $7$ which is coming $pH$ range between $6and8.$