Question

A 0.4322 g sample of a potassium hydroxide – lithium hydroxide mixture requires 27.10 mL of 0.3565 M HCl for its titration to the equivalence point. Calculate the mass percent lithium hydroxide in the mixture.

Answer 1

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:  

$\% = \frac{m_{LiOH}}{m_{t}} \times 100$    (1)

Where:

$m_{t} = m_{KOH} + m_{LiOH} = 0.4322 g$   (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.    

$\eta_{HCl} = \eta_{LiOH} + \eta_{KOH}$

$0.0271 L*0.3565 \frac{mol}{L} = \eta_{LiOH} + \eta_{KOH}$

$\eta_{LiOH} + \eta_{KOH} = 9.66 \cdot 10^{-3} \:mol$

Since mol = m/M, where M: is the molar mass and m is the mass, we have:

$\frac{m_{LiOH}}{M_{LiOH}} + \frac{m_{KOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol$    (3)                                        

Solving equation (2) for m_{KOH} and entering into equation (3), we can find the mass of LiOH:  

$\frac{m_{LiOH}}{M_{LiOH}} + \frac{0.4322 - m_{LiOH}}{M_{KOH}} = 9.66 \cdot 10^{-3} \:mol$    

$\frac{m_{LiOH}}{23.95 g/mol} + \frac{0.4322 g - m_{LiOH}}{56.1056 g/mol} = 9.66 \cdot 10^{-3} \:mol$              

Solving for $m_{LiOH}$, we have:

$m_{LiOH} = 0.082 g$

Hence, the percent lithium hydroxide is (eq 1):

$\% = \frac{0.082 g}{0.4322 g} \times 100 = 19.0 \%$  

Therefore, the mass percent lithium hydroxide in the mixture is 19.0%.

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I hope it helps you!