Question

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

Answer 1

`0.00833M`

Explanation:

We're asked to find the molar concentration of the `"NaCl"` solution given some titration data.

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

`"NaOH"(aq) + "HCl" (aq) rarr "NaCl" (aq) + "H"_2"O" (l)`

Using the molarity equation, we can find the number of moles of `"HCl"` that reacted:

`"molarity" = "mol solute"/"L soln"`

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

`"mol HCl" = (0.105"mol"/(cancel("L")))(0.0250cancel("L")) = 0.00263` `"mol HCl"`

(volume converted to liters)

Now, using the coefficients of the chemical reaction, we can determine the number of moles of `"NaOH"` that reacted:

`0.00263cancel("mol HCl")((1color(white)(l)"mol NaOH")/(1cancel("mol HCl"))) = 0.00263` `"mol NaOH"`

Lastly, we'll use the molarity equation (using given volume of `"NaOH soln"`) again to determine the molarity of the sodium hydroxide solution:

`"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)