In a titration of 250.0 mL of methanoic acid ("HCHO"_2), takes 16.32 mL of 0.1004 mol/L NaOH to reach the end-point. What is the molar concentration of the methanoic acid?

Jan 2, 2017

The concentration of the methanoic acid is 0.006 556 mol/L.

Explanation:

A titration calculation is really a stoichiometry problem.

You always use the same four steps.

Step 1. Write the balanced equation

$\text{HCOOH + NaOH" → "HCOONa" + "H"_2"O}$

Step 2. Calculate the moles of NaOH

$\text{Moles of NaOH" = "0.016 32" color(red)(cancel(color(black)("L NaOH"))) × "0.1004 mol NaOH"/(1 color(red)(cancel(color(black)("L NaOH")))) = "0.001 639 mol NaOH}$

Step 3. Calculate the moles of $\text{HCOOH}$

$\text{Moles of HCOOH" = "0.001 639" color(red)(cancel(color(black)("mol NaOH"))) × "1 mol HCOOH"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.001 639 mol HCOOH}$

Step 4. Calculate the molarity of the $\text{HCOOH}$

$\text{Molarity of HCOOH" = "moles"/"litres" = "0.001 639 mol"/"0.250 00 L" = "0.006 556 mol/L}$

(Note: I think the volume should have been 25.00 mL.)