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

Answer:

12.8

Explanation:

Considering:

$Molarity=\frac{Mass}{Molar mass\times Volume\ of\ the\ solution}$

For NaCl:-

Mass = 12.25 g

Molar mass = 58.44 g/mol

Volume of solution = 250.0 mL = 0.25 L

So,

$Molarity_{stock\ solution}=\frac{12.25}{39.997\times 0.25}\ M=1.225\ M$

Considering

$Molarity_{working\ solution}\times Volume_{working\ solution}=Molarity_{stock\ solution}\times Volume_{stock\ solution}$

Given  that:

$Molarity_{working\ solution}=?$

$Volume_{working\ solution}=1.00\ L$

$Volume_{stock\ solution}=0.05\ L$

$Molarity_{stock\ solution}=1.225\ M$

So,  

$Molarity_{working\ solution}\times 1.00=1.225\times 0.05$

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