Question

An analytical chemist is titrating of a solution of cyanic acid with a solution of . The of cyanic acid is . Calculate the pH of the acid solution after the chemist has added of the solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of solution added.

Answer 1

Answer: The pH of the solution is 12.73

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 cyanic acid:

Molarity of cyanic acid = 0.8600 M (Assuming)

Volume of solution = 179.5 mL (Assuming)

Putting values in equation 1, we get:

`0.8600M=\frac{\text{Moles of cyanic acid}* 1000}{179.5mL}\\\\\text{Moles of cyanic acid}=(0.8600* 179.5)/(1000)=0.15437mol`

• For NaOH:

Molarity of NaOH = 0.6500 M (Assuming)

Volume of solution = 274.6 mL (Assuming)

Putting values in equation 1, we get:

`0.6500M=\frac{\text{Moles of NaOH}* 1000}{274.6mL}\\\\\text{Moles of NaOH}=(0.6500* 274.6)/(1000)=0.17849mol`

Calculating the remaining moles of NaOH, we get:

`\text{Remaining moles of NaOH}=\text{Moles of NaOH - Moles of cyanic acid}\\\\\text{Remaining moles of NaOH}=0.17849-0.15437=0.02412mol`

Calculating the concentration of NaOH by using equation 1, we get:

Moles of NaOH = 0.02412 moles

Volume of solution = 179.5 + 274.6 = 454.1 mL

Putting values in above equation, we get:

`\text{Molarity of NaOH}=(0.02412* 1000)/(454.1)\\\\\text{Molarity of NaOH}=0.0531M`

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^-]=0.0531M`

Putting values in above equation, we get:

`pOH=-\log(0.0531)\\\\pOH=1.27`

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

`pH+pOH=14\\pH=14-1.27=12.73`

Hence, the pH of the solution is 12.73