What is the molarity of HCl found in a titration where 30 ml of HCl is titrated with 15 ml of .1M NaOH?

2 Answers
Jun 4, 2018

NaOH(aq) + HCl(aq) -> NaCl(aq) + H_2O(l)

The Molarity of HCl is .05M

Explanation:

Since in the balanced equation, NaOH and HCl have a 1:1 ratio they must have the same number of mole

We can say that mols is equal to (Molarity)/(Liters) since the units cancel out to equal moles ((mol)/(L))/(L) or (mol * L)/L or mol
Finally, we can set up a ratio to solve for the molarity of HCl

(.1_"M")/(15_"ml") = (x_"M")/(30_"ml")

This comes out to x=.05_"M" or HCl=.05_"M"

Jun 5, 2018

I get "0.05 M", which is a reasonable answer.


Write the chemical reaction. This neutralization is of a strong base and strong acid, and is a double-replacement reaction:

"NaOH"(aq) + "HCl"(aq) -> "H"_2"O"(l) + "NaCl"(aq)

This is 1:1, i.e. "1 mol OH"^(-) reacted with "1 mol H"^(+). Therefore, what we are looking for is for equal mols of "HCl" to react with "NaOH".

"0.1 mol NaOH"/cancel"L" xx 15 "m"cancel"L" = "1.5 mmols NaOH"

= "1.5 mmols OH"^(-)

This reacts exactly with "1.5 mmols H"^(+), or "1.5 mmols HCl".

If this is contained in "30 mL" of aqueous "HCl" solution, then the molar concentration is:

color(blue)(["HCl"]) = (1.5 cancel"m""mols HCl")/(30 cancel"m""L solution") = color(blue)("0.05 M HCl")