Question

A student prepares \(455 \mathrm{~mL}\) of a KOH solution, but neglects to write down the mass of \(\mathrm{KOH}\) added. His TA suggests that he take the \(\mathrm{pH}\) of the solution. The \(\mathrm{pH}\) is \(13.33 .\) How many grams of KOH were added?

Short answer

Answer: The student added approximately 5.47 grams of KOH to the solution.

Step-by-step solution

Calculate the pOH of the solution.

To find the pOH, we will use the following relationship between pH and pOH: pH + pOH = 14 Given pH = 13.33, we can now calculate the pOH: pOH = 14 - pH = 14 - 13.33 = 0.67

Determine the concentration of hydroxide ions (OH-).

The pOH relates to the concentration of hydroxide ions (OH-) through the formula: pOH = -log[OH-] Now we will use the pOH value to calculate the [OH-]: [OH-] = 10^(-pOH) = 10^(-0.67) mol/L ≈ 0.215 mol/L

Find the moles of KOH.

Since the concentration of hydroxide ions (OH-) is directly related to the concentration of KOH, we can use the [OH-] to find the moles of KOH in the solution. We will use the volume of the solution (455 mL) and the concentration ([OH-] = 0.215 mol/L) to calculate the moles of KOH: moles of KOH = [OH-] × volume moles of KOH = 0.215 mol/L × 455 mL × (1 L/1000 mL) ≈ 0.0978 mol

Calculate the mass of KOH.

To find the mass of KOH, we will use the molar mass of KOH (M(K) = 39 g/mol, M(O) = 16 g/mol, M(H) = 1 g/mol): molar mass of KOH = 39 + 16 + 1 = 56 g/mol Now, we will multiply the moles of KOH by its molar mass to obtain the mass of KOH: mass of KOH = moles of KOH × molar mass of KOH mass of KOH = 0.0978 mol × 56 g/mol ≈ 5.47 g Therefore, the student added approximately 5.47 grams of KOH to the solution.