a 20.0ml sample of HCL solution is placed in a flask with a few drops of indicator. If 30.0ml of a 0.240 M NaOH solution is needed to reach the endpoint, what is the molarity of the HCL solution?
mLa x Ma = mLb x Mb
To determine the molarity of the HCl solution, we can use a simple stoichiometry and dilution calculation. Here's how you can solve it:
1. The balanced equation for the neutralization reaction between HCl and NaOH is:
HCl + NaOH → H2O + NaCl
This equation tells us that 1 mole of HCl reacts with 1 mole of NaOH to give 1 mole of water and 1 mole of NaCl.
2. From the given information, we know that it took 30.0 ml of a 0.240 M NaOH solution to reach the endpoint. This means that 0.240 moles of NaOH were required to react with the HCl.
3. Now, we need to determine the number of moles of HCl in the 20.0 ml sample.
To do this, we can set up a mole-to-mole ratio using the balanced equation:
1 mole of HCl → 1 mole of NaOH
Since we already know that 0.240 moles of NaOH reacted, we can conclude that 0.240 moles of HCl must have been present in the 20.0 ml sample.
4. Now, we calculate the molarity of the HCl solution:
Molarity (M) = moles of solute / volume of solution (in liters)
We know that the volume of the HCl solution is 20.0 ml, which is equivalent to 0.0200 liters.
So, the molarity of the HCl solution can be calculated as:
Molarity = 0.240 moles / 0.0200 liters
Molarity = 12.0 M
Therefore, the molarity of the HCl solution is 12.0 M.