# A 25.0 mL sample of 0.105 M HCl was titrated with 315 mL of NaOH. What is the concentration of the NaOH?

Jul 10, 2017

$0.00833 M$

#### Explanation:

We're asked to find the molar concentration of the $\text{NaCl}$ solution given some titration data.

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

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

Using the molarity equation, we can find the number of moles of $\text{HCl}$ that reacted:

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

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

"mol HCl" = (0.105"mol"/(cancel("L")))(0.0250cancel("L")) = 0.00263 $\text{mol HCl}$

(volume converted to liters)

Now, using the coefficients of the chemical reaction, we can determine the number of moles of $\text{NaOH}$ that reacted:

0.00263cancel("mol HCl")((1color(white)(l)"mol NaOH")/(1cancel("mol HCl"))) = 0.00263 $\text{mol NaOH}$

Lastly, we'll use the molarity equation (using given volume of $\text{NaOH soln}$) again to determine the molarity of the sodium hydroxide solution:

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

M_ "NaOH" = (0.00263color(white)(l)"mol")/(0.315color(white)(l)"L") = color(red)(0.00833M

(volume converted to liters)