Question

What mass of \(\mathrm{KOH}\) is necessary to prepare \(800.0 \mathrm{~mL}\) of a solution having a \(\mathrm{pH}=11.56\) ?

Short answer

Approximately 0.161 g of \(\mathrm{KOH}\) is necessary to prepare 800.0 mL of the solution having a \(\mathrm{pH}=11.56\).

Step-by-step solution

Convert pH to pOH

: Since we're working with \(\mathrm{KOH}\), we need to find the pOH of the solution before we can find the concentration of hydroxide ions. To do this, we will use the relation: \( pOH = 14 - pH \) \( pOH = 14 - 11.56\) \( pOH = 2.44 \)

Calculate the concentration of \(\mathrm{OH^-}\) ions

: Now we have the pOH, we can find the concentration of \(\mathrm{OH^-}\) ions in the solution using the formula pOH = -log\(_{10}\)(\[\mathrm{OH^-}\]), where [\[\mathrm{OH^-}\]] denotes the concentration of \(\mathrm{OH^-}\) ions. Rearrange the formula and solve for [\[\mathrm{OH^-}\]] using the calculated pOH: \[ \mathrm{OH^{-}} = 10^{-pOH} \] \[ \mathrm{OH^{-}} = 10^{-2.44} \] \[ \mathrm{OH^{-}} ≈ 3.67 × 10^{-3} \mathrm{M} \]

Calculate moles of \(\mathrm{KOH}\)

: Since 1 mole of \(\mathrm{KOH}\) dissociates into 1 mole of \(\mathrm{OH^-}\) ions, we can find the moles of \(\mathrm{KOH}\) by multiplying the concentration by the volume of the solution. Moles of \(\mathrm{KOH}\) = Molarity × Volume Moles of \(\mathrm{KOH}\) = \((3.67 × 10^{-3}) \mathrm{M} × 0.800 \mathrm{L}\) (Remember to convert the 800.0 mL to L) Moles of \(\mathrm{KOH}\) ≈ 0.00293 mol

Calculate mass of \(\mathrm{KOH}\)

: Finally, we need to convert the moles of \(\mathrm{KOH}\) to grams using the molar mass (MM) of \(\mathrm{KOH}\) (39.10 g/mol for K, 15.999 g/mol for O, and 1.008 g/mol for H). Mass of \(\mathrm{KOH}\) = moles × molar mass Mass of \(\mathrm{KOH}\) = \(0.00293 \mathrm{mol} × (39.10 \mathrm{g/mol} + 15.999 \mathrm{g/mol} + 1.008 \mathrm{g/mol})\) Mass of \(\mathrm{KOH}\) ≈ 0.161 g Hence, approximately 0.161 g of \(\mathrm{KOH}\) is necessary to prepare 800.0 mL of the solution having a \(\mathrm{pH}=11.56\).