Question

Repeat the procedure in Exercise 67, but for the titration of 25.0 mL of 0.100 M HNO₃ with 0.100 M NaOH.

Short answer

For this case, pH will be 7.

Step-by-step solution

Finding pH by addition of 4 ml base 

The concentration of hydrogen ions are$\frac{2.1\times {10}^{-3}\mathrm{mol}}{(25+4)\times {10}^{-3}\mathrm{L}}=0.072 \mathrm{mol}/\mathrm{L}$

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

Finding pH by addition of 8 ml base 

The concentration of hydrogen ions are $\frac{1.7\times {10}^{-3}\mathrm{mol}}{(25+8)\times {10}^{-3}\mathrm{L}}=0.051 \mathrm{mol}/\mathrm{L}$The pH of the solution is$-\log [0.051]=1.28$

Finding pH by addition of 25 ml base 

The neutralization for total amount of acid are$0.1 \mathrm{mol}/\mathrm{L}\times 25\times {10}^{-3}\mathrm{L}=2.5\times {10}^{-3}\mathrm{mol}$ The amount of base added are $0.1 \mathrm{mol}/\mathrm{L}\times 25\times {10}^{-3}\mathrm{L}=25\times {10}^{-4}\mathrm{mol}$Thus, the amount of acid remains in the solution are The pH of the solution is 7.

Finding pH by addition of 30 ml base

The concentration of hydrogen ions are $\frac{0.5\times {10}^{-3}\mathrm{mol}}{(25+30)\times {10}^{-3}\mathrm{L}}=0.009 \mathrm{mol}/\mathrm{L}$The pH of the solution can be calculated as,

$\mathrm{pH}=14-2.24=11.95$