Can someone please explain me how to do this problem?
A 25mL solution of 0.5 M NaOH is titrated until neutralization into a 50mL sample of HCl. What was the concentration of the HCl?
I just got the answer... it's 0.25 M right?
mL x M = mL x M
25 mL x 0.5M = 50 mL x M
M HCl = (25 x 0.5/50) = 0.25M is right although the number of significant figures is not right IF 0.5, 25, and 50 are the actual numbers. I suspect you may have omitted the zeros on 0.50 etc.
To find the concentration of HCl, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between NaOH and HCl. The balanced equation is:
NaOH + HCl -> NaCl + H2O
The ratio of moles between NaOH and HCl is 1:1. Therefore, if we can determine the number of moles of NaOH used in the reaction, we will know the number of moles of HCl present.
To find the number of moles of NaOH, we can use the formula:
moles = concentration × volume
Given that the concentration of NaOH is 0.5 M and the volume is 25 mL (or 0.025 L), we can calculate the moles of NaOH:
moles of NaOH = 0.5 mol/L × 0.025 L = 0.0125 mol
Since the reaction is 1:1, this means that there are also 0.0125 mol of HCl present.
Now, to find the concentration of HCl, we can use the formula:
concentration = moles / volume
The volume of HCl is given as 50 mL (or 0.05 L), so we can substitute the values into the formula:
concentration of HCl = 0.0125 mol / 0.05 L = 0.25 M
Therefore, the concentration of the HCl solution is 0.25 M.