Chapter 8, Problem 144AE

Interpretation Introduction

Interpretation:

The pH values after the addition of each proportion of the base to the acid is to be determined. The titration curve needs to be drawn.

Concept introduction:

A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Sodium hydroxide is a strong base while nitric acid is a strong acid. When these two chemical species are reacted with each other, they form their ionic forms and a neutral solution.

Explanation of Solution

Initial pH of the analyte solution;

Nitric acid is a strong acid that completely dissociates in water.

The amount of hydrogen ion will be equal to the concentration of nitric acid thus,

  $\left[H^{+}\right]=0.1\text{ M}$

The pH of the solution will be:

  $pH=-\log [H^{+}]=-log[0.1]=1$

Addition of $4.0mL$of base:

Total amount of acid to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of base added $=0.100mol/L\times 4.0\times {10}^{-3}L=4\times {10}^{-4}mol$

Therefore, the amount of acid remaining in the solution $=(2.5-0.4)\times {10}^{-3}=2.1\times {10}^{-3}mol$

The concentration of hydrogen ions $=\frac{2.1\times {10}^{-3}mol}{(25.0+4.0)\times {10}^{-3}L}=0.072mol/L$

The pH of the solution $=-\log [0.072]=1.14$

Addition of $8.0mL$of base:

Total amount of acid to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of base added $=0.100mol/L\times 8.0\times {10}^{-3}L=8\times {10}^{-4}mol$

Therefore, the amount of acid remaining in the solution $=(2.5-0.8)\times {10}^{-3}=1.7\times {10}^{-3}mol$

The concentration of hydrogen ions $=\frac{1.7\times {10}^{-3}mol}{(25.0+8.0)\times {10}^{-3}L}=0.051mol/L$

The pH of the solution $=-\log [0.051]=1.28$

Addition of $12.5mL$of base:

Total amount of acid to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of base added $=0.100mol/L\times 12.5\times {10}^{-3}L=12.5\times {10}^{-4}mol$

Therefore, the amount of acid remaining in the solution $=(2.5-1.25)\times {10}^{-3}=1.25\times {10}^{-3}mol$

The concentration of hydrogen ions $=\frac{1.25\times {10}^{-3}mol}{(25.0+12.5)\times {10}^{-3}L}=0.033mol/L$

The pH of the solution $=-\log [0.033]=1.47$

Addition of $20.0mL$of base:

Total amount of acid to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of base added $=0.100mol/L\times 20.0\times {10}^{-3}L=20\times {10}^{-4}mol$

Therefore, the amount of acid remaining in the solution $=(2.5-2.0)\times {10}^{-3}=0.5\times {10}^{-3}mol$

The concentration of hydrogen ions $=\frac{0.5\times {10}^{-3}mol}{(25.0+20.0)\times {10}^{-3}L}=0.011mol/L$

The pH of the solution $=-\log [0.011]=1.95$

Addition of $24.0mL$of base:

Total amount of acid to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of base added $=0.100mol/L\times 24.0\times {10}^{-3}L=24\times {10}^{-4}mol$

Therefore, the amount of acid remaining in the solution $=(2.5-2.4)\times {10}^{-3}=0.1\times {10}^{-3}mol$

The concentration of hydrogen ions $=\frac{0.1\times {10}^{-3}mol}{(25.0+24.0)\times {10}^{-3}L}=0.002mol/L$

The pH of the solution $=-\log [0.002]=2.69$

Addition of $24.5mL$of base:

Total amount of acid to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of base added $=0.100mol/L\times 24.5\times {10}^{-3}L=24.5\times {10}^{-4}mol$

Therefore, the amount of acid remaining in the solution $=(2.5-2.45)\times {10}^{-3}=0.05\times {10}^{-3}mol$

The concentration of hydrogen ions $=\frac{0.05\times {10}^{-3}mol}{(25.0+24.5)\times {10}^{-3}L}=0.001mol/L$

The pH of the solution $=-\log [0.001]=2.99$

Addition of $24.9mL$of base:

Total amount of acid to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of base added $=0.100mol/L\times 24.9\times {10}^{-3}L=24.9\times {10}^{-4}mol$

Therefore, the amount of acid remaining in the solution $=(2.5-2.49)\times {10}^{-3}=0.01\times {10}^{-3}mol$

The concentration of hydrogen ions $=\frac{0.01\times {10}^{-3}mol}{(25.0+24.9)\times {10}^{-3}L}=0.0002mol/L$

The pH of the solution $=-\log [0.0002]=3.69$

Addition of $25.0mL$of base:

Total amount of acid to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of base added $=0.100mol/L\times 25.0\times {10}^{-3}L=25.0\times {10}^{-4}mol$

Therefore, the amount of acid remaining in the solution will be:

  $=(2.5-2.5)\times {10}^{-3}=0$ .

Since, both acid and base completely react to form salt and water, the pH of the solution $=7.00$

Addition of $28.0mL$of base:

Total amount of acid to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of base added $=0.100mol/L\times 28.0\times {10}^{-3}L=28\times {10}^{-4}mol$

Therefore, the amount of excess base in the solution $=(2.8-2.5)\times {10}^{-3}=0.3\times {10}^{-3}mol$

The concentration of hydroxide ions $=\frac{0.3\times {10}^{-3}mol}{(25.0+28.0)\times {10}^{-3}L}=0.0056mol/L$

Thenthe pH of the solution can be calculated as follows:

  $\begin{array}{l}pOH=-\log [0.0056]=2.24\\ pH=14.00-2.24=11.75\end{array}$

Addition of $30.0mL$of base:

Total amount of acid to be neutralized $=0.100mol/L\times 25.0\times {10}^{-3}L=2.5\times {10}^{-3}mol$

Amount of base added $=0.100mol/L\times 30.0\times {10}^{-3}L=30\times {10}^{-4}mol$

Therefore, the amount of excess base in the solution $=(3.0-2.5)\times {10}^{-3}=0.5\times {10}^{-3}mol$

The concentration of hydroxide ions $=\frac{0.5\times {10}^{-3}mol}{(25.0+30.0)\times {10}^{-3}L}=0.009mol/L$

Then the pH of the solution can be calculated as follows:

  $\begin{array}{l}pOH=-\log [0.009]=2.04\\ pH=14.00-2.24=11.95\end{array}$

The data obtained from the calculations is as follows:

Volume of base added (mL)	pH
0.0	1
4.0	1.14
8.0	1.28
12.5	1.47
20.0	1.95
24.0	2.69
24.5	2.99
24.9	3.69
25.0	7.0
28.0	11.75
30.0	11.95

The titration curve can be drawn as follows: