Question

An analytical chemist is titrating 242.5mL of a 1.200M solution of hydrazoic acid HN3 with a 0.3400M solution of NaOH . The pKa of hydrazoic acid is 4.72 . Calculate the pH of the acid solution after the chemist has added 1006.mL of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places.

Answer 1

Answer: The pH of the solution is 12.61

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$       ......(1)

• For hydrazoic acid:

Molarity of hydrazoic acid solution = 1.200 M

Volume of solution = 242.5 mL

Putting values in equation 1, we get:

$1.200M=\frac{\text{Moles of hydrazoic acid}\times 1000}{242.5mL}\\\\\text{Moles of hydrazoic acid}=0.291mol$

• For NaOH:

Molarity of NaOH solution = 0.3400 M

Volume of solution = 1006 mL

Putting values in equation 1, we get:

$0.3400M=\frac{\text{Moles of NaOH}\times 1000}{1006mL}\\\\\text{Moles of NaOH}=0.342mol$

The chemical reaction for hydrazoic acid and NaOH follows the equation:

                   $HN_3+NaOH\rightarrow NaN_3+H_2O$

Initial:           0.291        0.342

Final:                  0          0.051                 0.291      0.291

Volume of solution = 242.5 + 1006 = 1248.5 mL = 1.2485 L    (Conversion factor:  1 L = 1000 mL)

• For NaOH left:

Left moles of NaOH = 0.051 moles

Volume of the solution = 1.2485 L

Putting values in equation 1, we get:

$\text{Molarity of NaOH}=\frac{0.051mol}{1.2485L}=0.0408M$

1 mole of NaOH produces 1 mole of sodium ions and 1 mole of hydroxide ions

To calculate pOH of the solution, we use the equation:

$pOH=-\log[OH^-]$

We are given:

$[OH^-]=0.0408M$

Putting values in above equation, we get:

$pOH=-\log(0.0408)\\\\pOH=1.39$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\\\pH=14-1.39=12.61$

Hence, the pH of the solution is 12.61