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

The pH of the acid solution after adding NaOH can be calculated using the moles and concentrations of HN3 and NaOH involved in the titration. The final pH of the solution is found to be 12.61, indicating that the solution is basic.

Step-by-step explanation:

The pH of the acid solution after the chemist has added 1006 mL of the NaOH solution can be calculated by considering the moles of HN3 and NaOH involved in the reaction. The balanced chemical equation for the reaction is:

HN3 + NaOH → NaN3 + H2O

First, calculate the moles of HN3:

Moles of HN3 = volume (L) x concentration (M) = 0.2425 L x 1.200 M = 0.291 mol HN3

Then, calculate the moles of NaOH:

Moles of NaOH = volume (L) x concentration (M) = 1.006 L x 0.3400 M = 0.342 mol NaOH

Since the reaction is 1:1, the moles of HN3 and NaOH are equal at the equivalence point. This means that 0.291 mol of HN3 react with 0.291 mol of NaOH. The remaining moles of NaOH can be calculated by subtracting the moles of HN3 from the total moles of NaOH:

Moles of NaOH remaining = moles of NaOH - moles of HN3 = 0.342 mol - 0.291 mol = 0.051 mol NaOH

The volume of the final solution is the initial volume of the solution (242.5 mL) plus the volume of NaOH solution added (1006 mL), which is equal to 1.2485 L. To calculate the concentration of NaOH in the final solution, divide the moles of NaOH remaining by the volume of the final solution:

Concentration of NaOH in the final solution = moles of NaOH remaining / volume of final solution = 0.051 mol / 1.2485 L ≈ 0.0409 M

To calculate the pH of the final solution, we need to find the pOH first. The pOH can be calculated using the concentration of hydroxide ions ([OH-]) in the solution:

[OH-] = concentration of NaOH = 0.0409 M

pOH = -log([OH-]) = -log(0.0409) ≈ 1.39

Finally, calculate the pH using the equation:

pH = 14 - pOH = 14 - 1.39 = 12.61

Answer 2

Answer: The pH of the solution is 12.61

Step-by-step explanation:

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

`\text{Molarity of the solution}=\frac{\text{Moles of solute}* 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}* 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}* 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}=(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