Question

A solution of \(\mathrm{KOH}\) has a \(\mathrm{pH}\) of 13.29 . It requires \(27.66 \mathrm{~mL}\) of \(0.2500 \mathrm{MHCl}\) to reach the equivalence point. (a) What is the volume of the KOH solution? (b) What is the \(\mathrm{pH}\) at the equivalence point? (c) What is \(\left[\mathrm{K}^{+}\right]\) and \(\left[\mathrm{Cl}^{-}\right]\) at the equivalence point? Assume volumes are additive.

Short answer

Answer: (a) The volume of the KOH solution is 35.44 mL, (b) the pH at the equivalence point is 7, and (c) the concentrations of K+ and Cl- ions at the equivalence point are both 0.1095 M.

Step-by-step solution

Calculate the volume of the KOH solution

First, we need to determine the concentration of the KOH solution using its pH. The pH is 13.29, therefore the pOH can be calculated as: \(pOH = 14 - pH\) \(pOH = 14 - 13.29 = 0.71\) Now, we can find the concentration of the OH- ions in the KOH solution using the formula: \([OH^-] = 10^{-pOH} = 10^{-0.71} = 0.195 \text{ M}\) Since we know the volume of HCl needed to reach the equivalence point, we can use the mole-to-mole relationship to determine the volume of the KOH solution. At the equivalence point, the moles of OH- ions from KOH equals the moles of H+ ions from HCl. \(moles\ of\ HCl = moles\ of\ OH^-\) \(V_{HCl} × [HCl] = V_{KOH} × [OH^-]\) Now, plug in the given values and solve for \(V_{KOH}\): \(V_{KOH} = \frac{V_{HCl} × [HCl]}{[OH^-]} = \frac{27.66 \text{ mL} × 0.2500 \text{ M}}{0.195 \text{ M}} = 35.44 \text{ mL}\) (a) The volume of the KOH solution is 35.44 mL.

Determine the pH at the equivalence point

At the equivalence point, all the KOH reacts with the HCl to form water and KCl. Therefore, the pH at the equivalence point for a strong acid (HCl) and strong base (KOH) reaction will be 7. (b) The pH at the equivalence point is 7.

Find the concentrations of K+ and Cl- ions at the equivalence point

At the equivalence point, the concentrations of K+ and Cl- ions are equal and can be determined by considering the volume and concentration of both KOH and HCl solutions. Since we have the volume of both solutions and the HCl concentration, we can find the total volume of the mixture and the moles of Cl- ions produced by HCl: \(V_{total} = V_{HCl} + V_{KOH} = 27.66 \text{ mL} + 35.44 \text{ mL} = 63.1 \text{ mL}\) \(moles\ of\ Cl^- = V_{HCl} × [HCl] = 27.66 \text{ mL} × 0.2500 \text{ M} = 6.915 \text{ mmol}\) Now we can find the concentration of Cl- ions at the equivalence point: \([Cl^-] = \frac{moles\ of\ Cl^-}{V_{total}} = \frac{6.915 \text{ mmol}}{63.1 \text{ mL}} = 0.1095 \text{ M}\) Since the reaction produces equal amounts of K+ and Cl- ions, the concentration of K+ ions at the equivalence point is also 0.1095 M. (c) The concentration of K+ and Cl- ions at the equivalence point is 0.1095 M.