What is the HCl concentration if 65.8mL of 0.300M NaOH is required to titrate a 15.0mL sample of the acid?
mols HCl = M x L = ?
mols NaOH = the same
M NaOH = mols NaoH/L NaOH
(.0658*0.3)/(.015)
To find the concentration of HCl, we can use the equation:
M1V1 = M2V2
where M1 is the molarity of NaOH, V1 is the volume of NaOH used, M2 is the molarity of HCl, and V2 is the volume of HCl being titrated.
Given:
M1 (molarity of NaOH) = 0.300 M
V1 (volume of NaOH used) = 65.8 mL
V2 (volume of HCl being titrated) = 15.0 mL
First, we need to convert the volumes from milliliters (mL) to liters (L) to match the units in the molarity. We can do this by dividing the volumes by 1000:
V1 = 65.8 mL ÷ 1000 = 0.0658 L
V2 = 15.0 mL ÷ 1000 = 0.0150 L
Substituting the given values into the equation, we have:
(0.300 M)(0.0658 L) = M2(0.0150 L)
Now we can solve for M2, the molarity of HCl:
(0.300 M)(0.0658 L) ÷ (0.0150 L) = M2
0.01974 M = M2
Therefore, the concentration of HCl is 0.01974 M.