Suppose that we had used 25.5cm3 of 0.200M NaOH and 42cm3 of dilute HCl to reach the end point. What is the molarity of the Hcl acid?
HCl = NaOh--> NaCl + H20
Moles NaOH=.0255*.2=??
molarity of HCl= molesNaOH/.042
To find the molarity of the HCl acid, we can use the concept of stoichiometry.
First, let's compare the balanced chemical equation:
HCl + NaOH --> NaCl + H2O
From the equation, we can see that the molar ratio between HCl and NaOH is 1:1.
Given that you have used 25.5 cm3 of 0.200M NaOH and 42 cm3 of dilute HCl, we need to find the number of moles of NaOH used and use that as the number of moles of HCl.
Step 1: Calculate the number of moles of NaOH used.
Molarity (M) = moles (mol) / volume (L)
moles (mol) = Molarity (M) x volume (L)
= 0.200 mol/L x (25.5 cm3 / 1000 cm3/L)
= 0.0051 mol
Step 2: Since the molar ratio between HCl and NaOH is 1:1, the number of moles of HCl used is also 0.0051 mol.
Step 3: Calculate the molarity of HCl.
Molarity (M) = moles (mol) / volume (L)
Molarity (M) = 0.0051 mol / (42 cm3 / 1000 cm3/L)
= 0.1214 M
Therefore, the molarity of the HCl acid is 0.1214 M.