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..
To find the mass percent of lithium hydroxide in the mixture, we need to calculate the moles of lithium hydroxide and potassium hydroxide separately.
Let's start with the moles of HCl used in the titration. You have correctly calculated it as 0.010343 moles.
The balanced chemical equation between HCl and KOH is:
2HCl + KOH -> KCl + H2O
From the balanced equation, we can see that 2 moles of HCl react with 1 mole of KOH.
Since the molar ratio between HCl and KOH is 2:1, the number of moles of KOH can be calculated as half the moles of HCl, which is 0.010343/2 = 0.0051715 moles.
Now, to determine the moles of LiOH, we need to recognize that the total moles of hydroxide ions (OH-) in the mixture come from both KOH and LiOH.
The balanced chemical equation for LiOH reacting with HCl is:
LiOH + HCl -> LiCl + H2O
In this equation, we can see that one mole of LiOH reacts with one mole of HCl, which means the moles of LiOH are also equal to 0.010343.
Since we know the moles of LiOH and the total mass of the mixture, we can calculate the mass percent of LiOH in the mixture using the following formula:
Mass percent of LiOH = (moles of LiOH / total moles of mixture) x 100%
In this case, the total moles of the mixture are the sum of the moles of KOH and LiOH, which is 0.0051715 + 0.010343 = 0.0155145 moles.
Now, using the formula, the mass percent of LiOH = (0.010343 / 0.0155145) x 100% = 66.75%.
Therefore, the mass percent of lithium hydroxide in the mixture is approximately 66.75%.