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

Question

Peyton Adams · Accepted Answer

#0.00833M#

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)

Avery Adams · Answer

undefined The balanced chemical equation for the reaction between HCl and NaOH is:

$$ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} $$

Using the stoichiometry of the reaction, we find that one mole of HCl reacts with one mole of NaOH.

Given that 25.0 mL of 0.105 M HCl was titrated, we can calculate the number of moles of HCl:

$$ \text{moles of HCl} = \text{volume of HCl (L)} \times \text{concentration of HCl (M)} $$

$$ \text{moles of HCl} = 0.0250 \, \text{L} \times 0.105 \, \text{M} = 0.002625 \, \text{moles} $$

Since the stoichiometry is 1:1, the number of moles of NaOH used in the titration is also 0.002625 moles.

The concentration of NaOH can be calculated using the volume and number of moles of NaOH:

$$ \text{concentration of NaOH (M)} = \frac{\text{moles of NaOH}}{\text{volume of NaOH (L)}} $$

$$ \text{concentration of NaOH} = \frac{0.002625 \, \text{moles}}{0.315 \, \text{L}} $$

$$ \text{concentration of NaOH} \approx 0.00833 \, \text{M} $$

Therefore, the concentration of the NaOH solution is approximately 0.00833 M.