Chapter 18.3, Problem 33PP

(a)

Interpretation Introduction

Interpretation:

$k_a$ value of given acid is to be calculated.

Concept introduction:

Acid ionization constant tells about equilibrium that whether reactants are favored or products.

For a reaction,

$HA+H_{2}O\rightleftharpoons A^{-}+H_3{O}^{+}$

Acid ionization constant is given by:

$\begin{array}{l}{k}_a=\frac{\left[A^{-}\right]\times \left[H_3{O}^{+}\right]}{\left[HA\right]}\\ where,\\ \left[A^{-}\right]isconcentrationof{A}^{-}\\ \left[H_3{O}^{+}\right]isconcentrationof{H}_3{O}^{+}\\ \left[HA\right]isconcentrationofHA\end{array}$

And $pH+pOH=14$

Answer to Problem 33PP

$k_a=8.9\times {10}^{-5}$

Explanation of Solution

Ionization equation is given by:

$C_6{H}_{5}COOH+H_{2}O\rightleftharpoons C_6{H}_{5}CO{O}^{-}+H_3{O}^{+}$

Given data:

$Beforereaction,\left[C_6{H}_{5}COOH\right]=0.0033M$

$pOH=10.70$

$\begin{array}{c}pH=14-pOH\\ =14-10.70\\ =3.30\end{array}$

Now $\left[C_6{H}_{5}CO{O}^{-}\right]=\left[H^{+}\right]$

$pH=-\log \left[H^{+}\right]$

Therefore,

$\begin{array}{c}pH=3.30\\ -\log \left[H^{+}\right]=3.30\\ \left[H^{+}\right]=Anti\log (-3.30)\\ =5\times {10}^{-4}M\end{array}$

$\begin{array}{c}Atequilibrium,\left[C_6{H}_{5}COOH\right]=0.0033-5\times {10}^{-4}\\ =0.0028M\end{array}$

Therefore,

$\begin{array}{c}{k}_a=\frac{5\times {10}^{-4}\times 5\times {10}^{-4}}{3.5\times {10}^{-3}}\\ =8.9\times {10}^{-5}\end{array}$

(b)

Interpretation Introduction

Interpretation:

$k_a$ value of given acid is to be calculated.

Concept introduction:

Acid ionization constant tells about equilibrium that whether reactants are favored or products.

For a reaction,

$HA+H_{2}O\rightleftharpoons A^{-}+H_3{O}^{+}$

Acid ionization constant is given by:

$\begin{array}{l}{k}_a=\frac{\left[A^{-}\right]\times \left[H_3{O}^{+}\right]}{\left[HA\right]}\\ where,\\ \left[A^{-}\right]isconcentrationof{A}^{-}\\ \left[H_3{O}^{+}\right]isconcentrationof{H}_3{O}^{+}\\ \left[HA\right]isconcentrationofHA\end{array}$

And $pH+pOH=14$

Answer to Problem 33PP

$k_a=1\times {10}^{-5}$

Explanation of Solution

Ionization equation is given by:

$HCNO+H_{2}O\rightleftharpoons CN{O}^{-}+H_3{O}^{+}$

Given data:

$Beforereaction,\left[HCNO\right]=0.100M$

$pOH=11.00$

$\begin{array}{c}pH=14-pOH\\ =14-11\\ =3\end{array}$

Now $\left[CN{O}^{-}\right]=\left[H^{+}\right]$

$pH=-\log \left[H^{+}\right]$

Therefore,

$\begin{array}{c}pH=3\\ -\log \left[H^{+}\right]=3\\ \left[H^{+}\right]=Anti\log (-3)\\ ={10}^{-3}M\end{array}$

$\begin{array}{c}Atequilibrium,\left[HCNO\right]=0.1-1\times {10}^{-3}\\ =0.099M\end{array}$

Therefore,

$\begin{array}{c}{k}_a=\frac{1\times {10}^{-3}\times 1\times {10}^{-3}}{0.099}\\ =1\times {10}^{-5}\end{array}$

(c)

Interpretation Introduction

Interpretation:

$k_a$ value of given acid is to be calculated.

Concept introduction:

Acid ionization constant tells about equilibrium that whether reactants are favored or products.

For a reaction,

$HA+H_{2}O\rightleftharpoons A^{-}+H_3{O}^{+}$

Acid ionization constant is given by:

$\begin{array}{l}{k}_a=\frac{\left[A^{-}\right]\times \left[H_3{O}^{+}\right]}{\left[HA\right]}\\ where,\\ \left[A^{-}\right]isconcentrationof{A}^{-}\\ \left[H_3{O}^{+}\right]isconcentrationof{H}_3{O}^{+}\\ \left[HA\right]isconcentrationofHA\end{array}$

And $pH+pOH=14$

Answer to Problem 33PP

$k_a=1.5\times {10}^{-5}$

Explanation of Solution

Ionization equation is given by:

$C_3{H}_{7}COOH+H_{2}O\rightleftharpoons C_3{H}_{7}CO{O}^{-}+H_3{O}^{+}$

Given data:

$Beforereaction,\left[C_3{H}_{7}COOH\right]=0.150M$

$pOH=11.18$

$\begin{array}{c}pH=14-pOH\\ =14-11.18\\ =2.82\end{array}$

Now $\left[C_3{H}_{7}CO{O}^{-}\right]=\left[H^{+}\right]$

$pH=-\log \left[H^{+}\right]$

Therefore,

$\begin{array}{c}pH=2.82\\ -\log \left[H^{+}\right]=2.82\\ \left[H^{+}\right]=Anti\log (-2.82)\\ =1.5\times {10}^{-3}M\end{array}$

$\begin{array}{c}Atequilibrium,\left[C_3{H}_{7}COOH\right]=0.150-1.5\times {10}^{-3}\\ =0.149M\end{array}$

Therefore,

$\begin{array}{c}{k}_a=\frac{1.5\times {10}^{-3}\times 1.5\times {10}^{-3}}{0.149}\\ =1.5\times {10}^{-5}\end{array}$