The mass percent lithium hydroxide in the mixture with potassium hydroxide, calculated from the equivalence point in the titration of HCl with the mixture, is 19.0%.
The mass percent of lithium hydroxide can be calculated with the following equation:
(1)
Where:
(2)
We need to find the mass of LiOH.
From the titration, we can find the number of moles of the mixture since the number of moles of the acid is equal to the number of moles of the bases at the equivalence point.
![\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}](https://tex.z-dn.net/?f=%20%5Ceta_%7BHCl%7D%20%3D%20%5Ceta_%7BLiOH%7D%20%2B%20%5Ceta_%7BKOH%7D%20)
![0.0271 L*0.3565 \frac{mol}{L} = \eta_{LiOH} + \eta_{KOH}](https://tex.z-dn.net/?f=%200.0271%20L%2A0.3565%20%5Cfrac%7Bmol%7D%7BL%7D%20%3D%20%5Ceta_%7BLiOH%7D%20%2B%20%5Ceta_%7BKOH%7D%20)
![\eta_{LiOH} + \eta_{KOH} = 9.66 \cdot 10^{-3} \:mol](https://tex.z-dn.net/?f=%20%5Ceta_%7BLiOH%7D%20%2B%20%5Ceta_%7BKOH%7D%20%3D%209.66%20%5Ccdot%2010%5E%7B-3%7D%20%5C%3Amol%20)
Since mol = m/M, where M: is the molar mass and m is the mass, we have:
(3)
Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:
Solving for
, we have:
![m_{LiOH} = 0.082 g](https://tex.z-dn.net/?f=%20m_%7BLiOH%7D%20%3D%200.082%20g%20)
Hence, the percent lithium hydroxide is (eq 1):
Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.
Learn more about mass percent here:
I hope it helps you!
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