i did an experiment where we titrated HCl on calcium hydroxide, and now i am confused with the calculations, i do not know if i am correct, so please help me out...
Trial One
volume of HCl used - 11.20mL
concentration of HCl - 0.04mol/L
so to get moles of HCl i used M= CV
is that correct?
than to get moles of OH- in the solution, i did (M*V)/ 10mL(vlume of calcium hydroxide)
is that correct?
but then how do i find out the concentration of [OH-] and [Ca2+] at equilibrium. when i a supposed to even find out the Ksp int he same experiment?
Your information is sketchy at best. For example, where did the 10 mL come from?
You don't know what you have if you don't write the equation of what's happening.
Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O
moles HCl = M x L = ??
Using the coefficients in the balanced equation, convert moles HCl to moles Ca(OH)2. Therefore, moles Ca(OH)2 = 1/2 the moles HCl. Then since there are two OH^- in 1 molecule Ca(OH)2, there must be 2xmoles Ca(OH)2 = (OH^-).
So now you have Ca^2+ and OH^-, provided the 10 mL doesn't mess things around.
To calculate moles of HCl used in the experiment, you correctly used the formula M = CV, where M is the molarity, C is the concentration, and V is the volume. So, the moles of HCl used can be calculated as:
Moles of HCl = (Concentration of HCl) * (Volume of HCl used)
= (0.04 mol/L) * (11.20 mL)
= 0.448 moles
Next, to determine the moles of OH- in the solution, you need to consider the balanced chemical equation for the reaction between HCl and Ca(OH)2:
2HCl + Ca(OH)2 -> CaCl2 + 2H2O
From the balanced equation, you can see that for every 2 moles of HCl, 1 mole of Ca(OH)2 reacts. Therefore, the moles of OH- can be calculated as:
Moles of OH- = (Moles of HCl) / 2
= 0.448 moles / 2
= 0.224 moles
Now, to find the concentration of [OH-] and [Ca2+] at equilibrium, you need to consider the stoichiometry of the reaction. Since the reaction between HCl and Ca(OH)2 is a 1:2 ratio, for every mole of Ca(OH)2 that reacts, 2 moles of OH- and 1 mole of Ca2+ are formed.
Therefore, the concentration of OH- at equilibrium can be calculated as:
[OH-] at equilibrium = (Moles of OH-) / (Volume of Ca(OH)2 solution)
= 0.224 moles / 10 mL
= 0.0224 mol/L
Similarly, the concentration of Ca2+ at equilibrium would be:
[Ca2+] at equilibrium = (Moles of Ca2+) / (Volume of Ca(OH)2 solution)
= (0.224 moles * 1) / 10 mL
= 0.0224 mol/L
To determine the Ksp (solubility product constant) in the same experiment, you would need additional information, such as the balanced equation for the dissociation of Ca(OH)2 and the concentrations of the ions at equilibrium. The Ksp is calculated using the concentrations of the ions raised to their respective stoichiometric coefficients.
Without that information, it is not possible to calculate the Ksp value from the given data alone.