0.4276-g sample of a potassium hydroxide – lithium hydroxide mixture requires 31.20 mL of 0.3315 M HCl for its titration to the equivalence point. What is the mass percent lithium hydroxide in this mixture?
1.) 0.03120 L x 0.3315= 0.010343 moles HCl
I'm not sure where I go from there..
@DrBob22 thank you for explaining. But how would you solve for x?
(x/23.947)+(y/56.105)=0.01034
(X/23.947))=0.01034-(y/56.105)...
To find the mass percent of lithium hydroxide in the mixture, you need to calculate the moles of lithium hydroxide and the total mass of the mixture.
2.) First, let's calculate the moles of hydrochloric acid (HCl) reacted in the titration:
We already have the volume of HCl used (0.03120 L) and the molarity of HCl (0.3315 M). Multiply these two values together to get the moles of HCl used in the reaction:
0.03120 L x 0.3315 M = 0.010343 moles HCl
3.) Next, we need to determine the stoichiometry of the reaction between HCl and the mixture. From the balanced equation of the reaction, we can see that the ratio between HCl and lithium hydroxide (LiOH) is 1:1. This means that for every mole of HCl used, there is also one mole of LiOH in the mixture.
4.) Since the moles of HCl reacted is equal to the moles of LiOH in the mixture, we can say that there are 0.010343 moles of LiOH in the mixture.
5.) Finally, to calculate the mass percent of LiOH in the mixture, divide the mass of LiOH by the mass of the entire mixture and multiply by 100.
Mass percent of LiOH = (mass of LiOH / mass of mixture) x 100
Unfortunately, we don't have enough information to directly determine the mass of the mixture, as we only know the mass of the sample (0.4276 g) without distinguishing between potassium hydroxide (KOH) and LiOH.
This is not an easy problem. You set up two equations and solve them simultaneously. I will give you the two equations and tell you what to do after that.
Let X = grams LiOH
and Y = grams KOH
equation 1 is X+Y=0.4276
Equation 2 is obtained by an equation that adds mols LiOH (in terms of X) + mols KOH (in terms of Y) = total mols HCl I will use mm for molar mass.
(X/mm LiOH) + (Y/mm KOH) = total mols HCl = M HCl x L HCl.
Solve these equations simultaneously for X (and Y if you want it).
Then %LiOH = (grams LiOH/mass sample)*100 = ? Note: grams LiOH is X in your answer. Post your work if you get stuck.
Your problem isn't chemistry at this point; it's math. You haven't used the first equation of X+Y = 0.4276. Solve that for Y as
Y = 0.4276-X and substitute that into the equation you have for Y. That will give you just one unknown of X and you solve for that.