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

1 Answer
Jul 10, 2017

Answer:

#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)