# If it takes 33.2 ml of 0.1 M NaOH to neutralize 40 ml of HCl, then what is the molarity of the hcl?

Jul 3, 2017

$0.083 M$

#### Explanation:

We're asked to find the molar concentration (molarity) of the $\text{HCl}$ solution, given some titration measurements.

Let's first write the chemical equation for this neutralization reaction:

$\text{NaOH"(aq) + "HCl" (aq) rarr "NaCl" (aq) + "H"_2"O} \left(l\right)$

Since we're given the molarity and volume of the $\text{NaOH}$ solution used, we can calculate the moles be using the molarity equation:

"mol solute" = ("molarity")("L soln")

= (0.1"mol"/(cancel("L")))(0.0332cancel("L")) = color(red)(0.00332 color(red)("mol NaOH"

Since all the coefficients in the chemical equation are $1$, the relative number of moles of $\text{HCl}$ used is also color(red)(0.00332 color(red)("mol".

Finally, let's use the molarity equation again to find the molarity of the $\text{HCl}$ solution (given the volume of $\text{NaOH soln}$ is $40$ $\text{mL}$, which must be in liters when using molarity equations):

$\text{molarity" = "mol solute"/"L soln}$

= (color(red)(0.00332)color(white)(l)color(red)("mol HCl"))/(0.040color(white)(l)"L") = color(blue)(0.083M