A 50mL solution of NaOH is neutralized by adding 37.5 mL of 0.045 M HCL. What was the original concentration of the NaOH solution?
Write the equation and balance it.
mols HCl = M x L = ?
Using the coefficients in the balanced equation, convert mols HCl to mols NaOH. Then, M NaOH = mols NaOH/L NaOH
What is the balanced equation? I cant figure out the answer
HCl is an acid.
NaOH is a base.
general equation is as follows:
an acid + a base = salt + water
HCl + NaOH ==> NaCl + H2O
To find the original concentration of the NaOH solution, we can use the concept of stoichiometry and the equation for the neutralization reaction between NaOH and HCl:
NaOH + HCl → NaCl + H2O
We are given the volume (50 mL) and concentration (0.045 M) of the HCl solution added to neutralize the NaOH. We need to determine the concentration of the NaOH solution.
Let's start by calculating the number of moles of HCl added to the solution:
moles of HCl = volume of HCl solution (L) x concentration of HCl (mol/L)
We need to convert the volume of HCl from mL to L:
volume of HCl = 37.5 mL ÷ 1000 mL/L = 0.0375 L
Substituting the values into the equation:
moles of HCl = 0.0375 L x 0.045 mol/L
moles of HCl = 0.0016875 mol
According to the balanced equation, the mole ratio between NaOH and HCl is 1:1. This means that the number of moles of NaOH is equal to the number of moles of HCl.
moles of NaOH = 0.0016875 mol
To find the concentration of the NaOH solution, we need to divide the moles of NaOH by the volume of the NaOH solution:
concentration of NaOH (mol/L) = moles of NaOH / volume of NaOH solution (L)
We are given the volume of the NaOH solution as 50 mL. Converting to liters:
volume of NaOH solution = 50 mL ÷ 1000 mL/L = 0.05 L
Substituting the values into the equation:
concentration of NaOH = 0.0016875 mol / 0.05 L
concentration of NaOH ≈ 0.03375 mol/L
Therefore, the original concentration of the NaOH solution was approximately 0.03375 M.