an unknown molarity of HCl acts as an analyte.50mL of it is placed into a conical flask ans a 0.1M of solution of NaOH was used as a reagent. The endpoint is pH=7, with a pKa of 6.5 chosen.The colour of the solution changed when 10mL of 0.1012M NaOH WAS ADDED..CALCULATE THE Number of moles of of HCl added during titration and its molarity
To calculate the number of moles of HCl added during titration and its molarity, we can use the concept of stoichiometry. Here's how you can do it step by step:
Step 1: Determine the number of moles of NaOH used:
Since we know the volume (10 mL = 0.01 L) and the molarity (0.1012 M) of NaOH added, we can calculate the number of moles using the formula:
moles = volume (L) x molarity (mol/L)
moles of NaOH = 0.01 L x 0.1012 mol/L = 0.001012 mol
Step 2: Use the stoichiometry of the balanced chemical equation to establish the mole ratio between HCl and NaOH:
The balanced equation for the reaction between HCl and NaOH is:
HCl + NaOH → NaCl + H2O
The stoichiometry tells us that the ratio between HCl and NaOH is 1:1.
Step 3: Determine the number of moles of HCl:
Since the mole ratio between HCl and NaOH is 1:1, the number of moles of HCl added is also 0.001012 mol.
Step 4: Calculate the molarity of HCl:
To find the molarity of HCl, we need to know the total volume of the HCl solution. However, the question only provides information about the volume of NaOH added (10 mL). We assume that no other NaOH was added prior to the endpoint condition (pH=7), meaning the HCl solution has reacted completely with the NaOH.
Since the endpoint condition (pH=7) is reached when an equal number of moles of HCl and NaOH have reacted, the volume of NaOH (10 mL) will be equal to the volume of HCl added. Therefore, the total volume of HCl solution is 10 mL.
Converting the volume to liters: 10 mL = 0.01 L
Now, we can calculate the molarity of HCl by dividing the moles of HCl by the volume in liters:
Molarity of HCl = moles of HCl / volume of HCl (L)
Molarity of HCl = 0.001012 mol / 0.01 L = 0.1012 M
Therefore, the number of moles of HCl added during titration is 0.001012 mol, and the molarity of HCl is approximately 0.1012 M.
To calculate the number of moles of HCl added during titration and its molarity, we need to use the concept of stoichiometry.
First, let's find the number of moles of NaOH used.
Moles of NaOH = Volume (in L) x Molarity
Moles of NaOH = 0.01 L x 0.1012 mol/L
Moles of NaOH = 0.001012 mol
Since the reaction between HCl and NaOH is a 1:1 ratio according to the balanced chemical equation, the number of moles of HCl added is also 0.001012 mol.
Next, we can calculate the molarity of the HCl solution.
Molarity = Moles of solute / Volume of solution (in L)
Molarity = 0.001012 mol / 0.05 L
Molarity = 0.02024 mol/L or 0.0202 M (rounded to 3 significant figures)
So, the number of moles of HCl added during titration is 0.001012 mol, and the molarity of the HCl solution is 0.0202 M.