A 0.4272-g sample of a potassium hydroxide – lithium hydroxide mixture requires 26.94mL of 0.3385M for its titration to the equivalence point.

What is the mass percent lithium hydroxide in this mixture?

To calculate the mass percent of lithium hydroxide in the mixture, we need to determine the mass of lithium hydroxide in the 0.4272-g sample.

Here's how you can calculate it step by step:

1. Calculate the number of moles of the titrant (the 0.3385M solution of an acid) used in the titration.
Moles of titrant = volume of titrant (L) × concentration of titrant (mol/L)
Moles of titrant = 26.94 mL × (1 L / 1000 mL) × 0.3385 mol/L

2. Determine the balanced chemical equation between the titrant and the mixture.
In this case, no specific chemical equation is given. However, we know that potassium hydroxide (KOH) and lithium hydroxide (LiOH) react with acids, so we can assume that the titrant reacts with both KOH and LiOH.

Let's assume the reaction is as follows:
HX + KOH → H2O + KX (1)
HX + LiOH → H2O + LX (2)

In each case, 1 mole of acid (HX) reacts with 1 mole of either KOH or LiOH, producing 1 mole of water and 1 mole of the corresponding salt (KX or LX).

3. Calculate the number of moles of lithium hydroxide in the titrated sample.
Since we assumed that the reaction between LiOH and the titrant is similar to equation (2), the number of moles of LiOH in the sample can be calculated by multiplying the moles of titrant by the stoichiometric ratio between LiOH and HX (1 mole of LiOH per mole of HX).

Moles of LiOH = moles of titrant × (1 mole of LiOH / 1 mole of HX)

4. Calculate the mass of LiOH in the sample.
Mass of LiOH = moles of LiOH × molar mass of LiOH

5. Calculate the mass percent of LiOH in the mixture.
Mass percent of LiOH = (mass of LiOH / mass of mixture) × 100

By following these steps, you can find the mass percent of lithium hydroxide in the given mixture.