Chapter 14, Problem 74QAP

Fifty cm³ of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution

(a) before any NaOH is added?

(b) at half-neutralization?

(c) at the equivalence point?

(d) when 0.10 mL less than the volume of NaOH to reach the equivalence point is added?

(e) when 0.10 mL more than the volume of NaOH to reach the equivalence point is added?

(f) Use your data to construct a plot similar to that shown in Figure 14.10 (pH versus volume NaOH added).

Interpretation Introduction

(a)

Interpretation:

The titration of NaOH of molarity 0.850 M occurs with nitrous acid of 50 cm³ with molarity 1.000M. The pH value of the solution is to be determined before the addition of NaOH.

Concept introduction:

For a reaction as follows:

$\text{AB}\rightleftharpoons \text{A}+\text{B}$

The equilibrium constant can be calculated as follows:

$K_a=\frac{[A][B]}{[AB]}$

The pH of the solution can be calculated as follows:

$pH=-{\log }_{10}[H^{+}]$

Answer to Problem 74QAP

The pH value of the solution is 1.62.

Explanation of Solution

The dissociation reaction for the nitrous acid is shown below-

${\text{HNO}}_2(aq)\to {\text{H}}^{+}(aq)+{\text{NO}}_2^{-}(aq)$

Given that-

Molarity of HNO₂ = 1.000M

The dissociation constant, Ka = 6.0×10-4

Consider the x M of HNO₂ is ionized. Hence the ICE table is shown below-

	HNO2	H+	NO2 -
I (mol)	1.000	0	0
C (mol)	-x	+x	+x
E (mol)	1.000 -x	x	x

Now the dissociation constant Ka is calculated as-

$K_a=\frac{[H^{+}][N{O}_2^{-}]}{[HN{O}_2]}$ (1)

Given that-

[HNO₂ ] = 1.000 -x

[H⁺] = x

[NO₂ ^-] = x

$K_a$ = 6.0×10-4

Put the above values in equation (1),

$6.0\times {10}^{-4}=\frac{x\times x}{1.000-x}$

On calculation −

x = 2.42×10-2

The concentration of [H⁺] = x= [NO₂ ^-] = 2.42×10-2 at equilibrium

Now, the pH value is −

$pH=-{\log }_{10}[H^{+}]$ ………………. (2)

$pH=-{\log }_{10}[2.42\times {10}^{-2}]$

The value of pH = 1.62

Interpretation Introduction

(b)

Interpretation:

The pH value of the solution is to be determined at half neutralization.

Concept introduction:

The $p{K}_a$ is negative log of the acid dissociation constant. Thus,

$p{K}_a=-{\log }_{10}{K}_a$

The pH of the solution can be calculated as follows:

$pH=-{\log }_{10}[H^{+}]$

Answer to Problem 74QAP

The pH value of the solution is 3.22 at half neutralization.

Explanation of Solution

At half neutralization, only half acid will be ionized and the equilibrium concentration of the conjugate base of acid and acid will be equal. Therefore,

$pH=p{K}_a$

$p{K}_a=-{\log }_{10}{K}_a$

Or

$pH=-{\log }_{10}{K}_a$.. .......(1)

And

K_a = 6.0×10-4

$pH=-{\log }_{10}6.0\times {10}^{-4}$

The value of pH for the solution = 3.22

Interpretation Introduction

(c)

Interpretation:

The pH value of the solution is to be determined at the equivalence point.

Concept introduction:

The molarity of solution is calculated as follows:

$\text{molarity=}\frac{\text{Number of moles}}{\text{volume}}$

Here, number of moles is of solute and volume is of solution.

For a base dissociation reaction,

$AB\rightleftharpoons A+B$

The base dissociation constant will be:

$K_b=\frac{[A][B]}{[AB]}$

It is related to acid dissocation constant and ionization constant of water as follows:

$K_b=\frac{K_w}{K_a}$

The pOH and pH of the solution can be calculated as follows:

$pOH=-{\log }_{10}[O{H}^{-}]$

$pH=-{\log }_{10}[H^{+}]$

Here,

$pH+pOH=14$

Answer to Problem 74QAP

The pH value of the solution is 8.45 at the equivalence point.

Explanation of Solution

The chemical equation for the reaction between nitrous acid and sodium hydroxide is given as-

${\text{HNO}}_{\text{2}}\text{(aq)+NaOH(aq)}\to {\text{H}}_{\text{2}}{\text{O(l)+Na}}^{\text{+}}{\text{(aq)+NO}}_{\text{2}}^{\text{-}}\text{(aq)}$

The net-ionic equation of the above reaction is −

${\text{HNO}}_{\text{2}}{\text{(aq)+OH}}^{\text{-}}\text{(aq)}\to {\text{NO}}_{\text{2}}^{\text{-}}{\text{(aq)+H}}_{\text{2}}\text{O(l)}$

Number of moles of nitrous acid-

$\text{molarity=}\frac{\text{Number of moles}}{\text{volume}}$

Or

$\text{Number of moles=molarity×volume}$ …......... (1)

Given that-

Molarity = 1.000M

Volume = 50.0cc = 0.050L

Put the above values in equation (1)

$\text{Number of moles=1}\text{.000M×0}\text{.050L}$

Number of moles of HNO₂ = 0.0500 mol

At equivalence point-

Number of moles of NaOH added = initial number of moles of HNO₂ = 0.0500 mol

Molarity of the base = 0.850 M

The volume of the base-

$\text{volume=}\frac{\text{Number of moles}}{\text{molarity}}$

Put the values in the above equation-

$\text{volume=}\frac{\text{0}\text{.0500mol}}{\text{0}\text{.850M}}$

Volume of base = 0.0588 L

Molarity of NO₂ - is calculated as-

Number of moles of HNO₂ = 0.0500mol = Number of moles of NO₂ -

Total volume = 0.0500L +0.0588L

$\text{molarity=}\frac{\text{Number of moles}}{\text{volume}}$

Put the given values in above equation-

$\text{molarity=}\frac{\text{0}\text{.0500mol}}{\text{0}\text{.0500L +0}\text{.0588L}}$

Molarity of NO₂ - = 0.459M

Consider the M of NO₂ - will be ionized. Hence the ICE table is shown below-

	NO2 -	HNO2	OH-
I (mol)	0.459	0	0
C (mol)	-y	+y	+y
E (mol)	0.459 -y	y	y

Now the base dissociation constant K_b is calculated as-

$K_b=\frac{[O{H}^{-}][HN{O}_2]}{[N{O}_2^{-}]}$ (2)

$K_b=\frac{K_w}{K_a}$ ………….. (2)

Given that-

[NO₂ ^-] = 0.459 −y

[HNO₂ ] = y

[OH^-] = y

$K_w$ = 1.0×10-14

$K_a$ = 6.0×10-4

Put the above values in equation (2)

$\frac{1.0\times {10}^{-14}}{6.0\times {10}^{-4}}=\frac{y\times y}{0.459-y}$

On calculation −

[OH^-] = y = 2.79×10-6

Now, the pH of the solution −

$pOH=-{\log }_{10}[O{H}^{-}]$.. ....... (3)

$pOH=-{\log }_{10}[2.79\times {10}^{-6}]$

$pOH=5.554$

And,

$pH+pOH=14$

$pH=14-5.554$

$pH=8.45$

Interpretation Introduction

(d)

Interpretation:

The volume less than the 0.10 mL of NaOH is added to reach the equivalence point. At this condition, the pH value of the solution is to be determined.

Concept introduction:

The molarity of solution is calculated as follows:

$\text{molarity=}\frac{\text{Number of moles}}{\text{volume}}$

Here, number of moles is of solute and volume is of solution.

Now, from the Henderson −Hasselbalch equation is as follows−

$pH=p{K}_a+\log \frac{[A^{-}]}{[HA]}$

Answer to Problem 74QAP

The pH value of the solution is 5.92, when the volume less than the 0.10mL of NaOH added.

Explanation of Solution

The number of moles of base-

$\text{Molarity=}\frac{\text{number of moles}}{\text{volume}}$.. .......(1)

Given that-

Volume = 58.7mL (0.1mL less than 58.8mL)

Molarity = 0.850 M

Put the above values in Equ (1)

$\text{number of moles=Molarity×volume}$

$\text{number of moles=0}\text{.850M×0}\text{.0587L}$

Number of moles of base (NaOH) = 0.0499mol

Now, the number of moles of HNO₂ −

$n_{HN{O}_2}=(0.0500-0.0499)$

$n_{HN{O}_2}=1.00\times {10}^{-4}mol$

Now, the number of moles of conjugate base, NO₂ -

$n_{N{O}_2^{-}}=0.0499$

Now, from the Henderson −Hasselbalch equation −

$\text{pH}=\text{p}{K}_a+\log \frac{[{\text{NO}}_2^{-}]}{[{\text{HNO}}_2]}$

Or,

$\text{pH}=3.22+\log \frac{\frac{\text{0}\text{.0499mol}}{\text{(50+58}\text{.7)mL}}}{\frac{\text{1}{\text{.00×10}}^{\text{-4}}\text{mol}}{\text{(50+58}\text{.7)mL}}}$

On calculation-

$\text{pH}=5.92$

Interpretation Introduction

(e)

Interpretation:

The volume more than the 0.10mL of NaOH is added to reach the equivalence point. At this condition, the pH value of the solution is to be determined.

Concept introduction:

The molarity of solution is calculated as follows:

$\text{molarity=}\frac{\text{Number of moles}}{\text{volume}}$

Here, number of moles is of solute and volume is of solution.

The pH of the solution can be calculated as follows:

$pH=-{\log }_{10}[H^{+}]$

Answer to Problem 74QAP

The pH value of the solution is 10.88 when the volume more than the 0.10mL of NaOH added.

Explanation of Solution

The number of moles of base-

$\text{Molarity=}\frac{\text{number of moles}}{\text{volume}}$.. .......(1)

Given that-

Volume = 58.9mL (0.1mL greater than 58.8mL)

Molarity = 0.850 M

Put the above values in equation (1)

$\text{number of moles=Molarity×volume}$

$\text{number of moles=0}\text{.850M×0}\text{.0589L}$

Number of moles of base (NaOH) = 0.050082mol

Now, the number of moles of OH- −

$n_{O{H}^{-}}=(0.050082-0.0500)$

$n_{O{H}^{-}}=8.2\times {10}^{-5}\text{mol}$

Now, pOH-

$pOH=-{\log }_{10}[O{H}^{-}]$.. .......(2) $\text{pOH}=-{\log }_{10}(8.2\times {10}^{-5}\text{mol})$

$pOH=3.12$

And

$\text{pH+pOH}=14$

$\text{pH}=14-3.12$

$\text{pH}=10.88$

Interpretation Introduction

(f)

Interpretation:

The graph is to be plotted by using the calculated data between pH vs. NaOH added.

Concept introduction:

The equivalence point is defined as the point for the titration process at which addition of titrant is enough for the complete neutralization of the analyte.

Half neutralization is defined as the point at which the concentration of weak acid will become equal to its conjugate base.

Explanation of Solution

The graph plot between pH vs. NaOH added is shown below-

According to the graph, the half neutralization point is approximately at pH 2.8 and equivalence point is approximately at pH 8.8