When a 0.5725 g sample of Lysol was titrated with 0.100 M NaOH, and endpoint was obtained at 15.00 mL. What is the percent of HCL in the lysol sample?
Technically, there may not be ANY HCl in the Lysol but you may calculate the acidity in terms of the HCl content.
moles NaOH = M x L = ??
moles HCl = same thing since 1 mole NaOH exactly neutralizes 1 mole HCl.
g HCl = moles HCl x molar mass HCl.
%HCl = (mass HCl/mass sample)&100 = ??
9.55%
To determine the percent of HCl in the Lysol sample, you will need to calculate the moles of HCl and the moles of Lysol in the sample.
First, let's calculate the moles of NaOH used in the titration. We know the volume and concentration of NaOH used:
moles of NaOH = concentration * volume
moles of NaOH = 0.100 M * 0.01500 L
Next, we need to determine the moles of HCl that reacted with the NaOH. According to the neutralization reaction between HCl and NaOH, they react in a 1:1 ratio. Thus, the moles of HCl will be the same as the moles of NaOH used:
moles of HCl = moles of NaOH
Now, since we have the mass of the Lysol sample, we can calculate the moles of Lysol using its molar mass (g/mol):
moles of Lysol = mass of Lysol / molar mass of Lysol
Lastly, divide the moles of HCl by the moles of Lysol and multiply by 100 to get the percentage:
percent of HCl = (moles of HCl / moles of Lysol) * 100
To finalize the calculation, you will need to know the molar mass of Lysol and HCl, which I will assume you have. Plug in the values into the equations, and you will have the percent of HCl in the Lysol sample.