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?
What did you titrate it with? It makes a difference. Also, did you type in all the question?
Sorry i guess i missed some. It is titrated with HCl
To find the mass percent of lithium hydroxide in the mixture, you need to determine the mass of lithium hydroxide in the 0.4272-g sample.
First, let's calculate the moles of potassium hydroxide used in the titration:
moles of potassium hydroxide = (volume of titrant) x (molarity of titrant)
= (26.94 mL) x (0.3385 mol/L)
= 0.009114 mol
Next, we need to determine the moles of hydroxide ions present in this amount of potassium hydroxide, as equal moles of hydroxide ions will be present in lithium hydroxide.
moles of hydroxide ions = moles of potassium hydroxide
= 0.009114 mol
Since lithium hydroxide has a 1:1 mole ratio with hydroxide ions, the moles of lithium hydroxide in the sample is also 0.009114 mol.
Now, we determine the molar mass of lithium hydroxide:
molar mass of lithium hydroxide = (molar mass of lithium) + (molar mass of oxygen) + (molar mass of hydrogen)
= (6.94 g/mol) + (16.0 g/mol) + (1.01 g/mol)
= 23.95 g/mol
To calculate the mass of lithium hydroxide in the sample, we can use the following formula:
mass of lithium hydroxide = (moles of lithium hydroxide) x (molar mass of lithium hydroxide)
= (0.009114 mol) x (23.95 g/mol)
= 0.2179 g
Finally, we can calculate the mass percent of lithium hydroxide in the mixture:
mass percent of lithium hydroxide = (mass of lithium hydroxide) / (mass of mixture) x 100%
= (0.2179 g) / (0.4272 g) x 100%
≈ 50.99%
Therefore, the mass percent of lithium hydroxide in the mixture is approximately 50.99%.