It is found that an average volume of 24.25 ml of approximately 0.1 M NaOH is required to reach an endpoint in a titration with 25.00ml of 0.09917 M HCl solution. What is the NaOH concentration? Make sure to give answer to four significant figures.
NaOH + HCl ==> NaCl + H2O
mols HCl = M x L = ?
mols NaOH = mols HCl (since the reaction shows 1 mol HCl = 1 mol NaOH
Then M NaOH = mols NaOH/L NaOH. The problem is confusing because it appears there is no unknown; ignore the "approximately 0.1 M NaOH". That's the unknown. That's what you're calculating in the last step.
To find the NaOH concentration, we can use the concept of stoichiometry and the balanced equation between NaOH and HCl.
The balanced equation for the reaction between NaOH and HCl is:
NaOH + HCl -> NaCl + H2O
From the balanced equation, we can see that the molar ratio between NaOH and HCl is 1:1. This means that every 1 mole of NaOH reacts with 1 mole of HCl.
Given that the volume of HCl used is 25.00 ml and the concentration of HCl is 0.09917 M, we can calculate the number of moles of HCl used.
moles of HCl = volume of HCl (in L) * concentration of HCl
= 25.00 ml * (1 L/1000 ml) * 0.09917 M
Now, using the molar ratio between NaOH and HCl of 1:1, we can say that the number of moles of HCl used is equal to the number of moles of NaOH used.
moles of NaOH = moles of HCl
Next, if the average volume of NaOH used is 24.25 ml and the concentration of NaOH is x M, we can calculate the number of moles of NaOH using the equation:
moles of NaOH = volume of NaOH (in L) * concentration of NaOH
= 24.25 ml * (1 L/1000 ml) * x M
Now, equating the two equations for moles of NaOH:
24.25 ml * (1 L/1000 ml) * x M = 25.00 ml * (1 L/1000 ml) * 0.09917 M
Simplifying the equation, we get:
0.02425 L * x M = 0.02500 L * 0.09917 M
Dividing both sides by 0.02425 L, we get:
x M = (0.02500 L * 0.09917 M) / 0.02425 L
Calculating the right side of the equation:
x M = 0.10249 M
Therefore, the concentration of NaOH is approximately 0.1025 M, rounded to four significant figures.