Question

Find the pH of a solution prepared by taking a 50.0 mL aliquot of a solution prepared by dissolving 12.25g of NaOH and 250.0 mL of water and diluting that aliquot to 1.00L?

Answer 1

12.8

Step-by-step explanation:

Considering:

`Molarity=(Mass)/(Molar mass* Volume\ of\ the\ solution)`

For NaCl:-

Mass = 12.25 g

Molar mass = 58.44 g/mol

Volume of solution = 250.0 mL = 0.25 L

So,

`Molarity_(stock\ solution)=(12.25)/(39.997* 0.25)\ M=1.225\ M`

Considering

`Molarity_(working\ solution)* Volume_(working\ solution)=Molarity_(stock\ solution)* Volume_(stock\ solution)`

Given that:

`Molarity_(working\ solution)=?`

`Volume_(working\ solution)=1.00\ L`

`Volume_(stock\ solution)=0.05\ L`

`Molarity_(stock\ solution)=1.225\ M`

So,

`Molarity_(working\ solution)* 1.00=1.225* 0.05`

Concentration of NaOH = Concentration of [OH⁻] = 0.06125 M

pOH = - log[OH⁻] = -log(0.06125) = 1.21

pH = 14 - pOH = 14 - 1.21 = 12.8