A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl
To calculate the amount of NaOH required to neutralize the HCl, we need to consider the reaction stoichiometry. The balanced chemical equation for the reaction between NaOH and HCl is:
NaOH + HCl -> NaCl + H2O
From the equation, we can see that 1 mole of NaOH reacts with 1 mole of HCl to produce 1 mole of NaCl and 1 mole of water.
First, let's calculate the number of moles of HCl in the 50 ml sample:
Moles of HCl = (concentration of HCl) * (volume of HCl in liters)
= (unknown concentration) * (50 ml / 1000 ml per liter)
= (unknown concentration) * 0.05 L
Since we don't know the concentration of HCl, we can use the concept of stoichiometry to determine the concentration of HCl from the known concentration of NaOH.
According to the reaction stoichiometry, the ratio of moles of HCl to NaOH is 1:1. Therefore, the moles of HCl and NaOH are equal.
Moles of HCl = Moles of NaOH
We can use the concentration and volume of NaOH solution to calculate the moles of NaOH:
Moles of NaOH = (concentration of NaOH) * (volume of NaOH in liters)
= 0.5 M * (25 ml / 1000 ml per liter)
= 0.5 M * 0.025 L
Since the moles of HCl and NaOH are equal, we can equate the two expressions:
(concentration of NaOH) * (volume of NaOH) = (unknown concentration) * (volume of HCl)
Plugging in the values, we have:
0.5 M * 0.025 L = (unknown concentration) * 0.05 L
Now we can solve for the unknown concentration:
(unknown concentration) = (0.5 M * 0.025 L) / 0.05 L
Simplifying the expression, we get:
(unknown concentration) = 0.25 M
Therefore, the concentration of the HCl in the 50 ml sample is 0.25 M.