How do i determine the exact molarity of .2 M of HCL?
The molarity of NaOH Used was .2010M used in the titration. And the volume used 10.33 mL used to reach equivalence point. What would be the Molarity of HCL?
It is unclear to me about the exact volumes and what is titrating what; however, here is how you do the problem.
Write and balance the equation.
HCl + NaOH ==> NaCl + H2O
moles NaOH = MNaOH x LNaOH = ??
Using the coefficients in the balanced equation, convert moles NaOH to moles HCl. It is 1:1; therefore, moles NaOH = moles HCl.
Then M HCl = molesHCl/LHCl.
To determine the exact molarity of HCl, you can use the concept of stoichiometry and the titration equation.
In this case, you have performed a titration where a known concentration of NaOH was used to neutralize the HCl. The balanced equation for the reaction is:
HCl + NaOH → NaCl + H2O
Based on the equation, the stoichiometric ratio is 1:1 between HCl and NaOH. This means that for every mole of HCl, you need one mole of NaOH to reach the equivalence point.
Given the volume of NaOH used (10.33 mL) and its molarity (0.2010 M), you can calculate the number of moles of NaOH used as follows:
moles of NaOH = (volume of NaOH used in liters) x (molarity of NaOH)
First, convert the volume from milliliters to liters:
volume of NaOH used = 10.33 mL = 10.33 x 10^(-3) L
Now, calculate the moles of NaOH used:
moles of NaOH = (10.33 x 10^(-3) L) x (0.2010 M)
Next, recall the stoichiometric ratio between HCl and NaOH (1:1). Since the stoichiometry is 1:1, the number of moles of HCl used would be the same as the moles of NaOH used.
Therefore, the molarity of HCl is equal to the moles of HCl divided by the volume of HCl used.
molarity of HCl = (moles of HCl) / (volume of HCl used in liters)
Since we haven't been given the volume of HCl used, we cannot determine the exact molarity of HCl with the information provided.