Suppose a chemistry problem asks for the percent of sodium hydroxide, NaOH, in a water solution. Assume 25.00 milliliters of the solution is titrated with a standard hydrochloric acid solution of concentration 0.129 molar HCL to an end-point indicated at 32.72 mL.
Note the molar concentration, or molarity (M), is the number of moles of a substance contained in a volume of 1 liter.
Write the balanced equation of the chemical reaction: NaOH + HCL = NaCL + H2O.
Note the molar ratio of the reaction is 1 mole NaOH / 1 mole HCL.
Determine the number of moles of HCL required to neutralize the NaOH, as follows:
32.72 ml HCL x (1 L / 1000 mL) x 0.129 moles HCL / L = 0.00442 moles HCL.
Calculate the molar concentration of NaOH, as follows: 0.00422 moles HCL x (1 mole NaOH / 1 mole HCL) / 25 mL x 1000 mL / 1 L = 0.169 M
Calculate the weight of NaOH per liter of solution by multiplying the molar concentration of NaOH by its molar mass, or formula weight, as follows: 0.169 moles NaOH / L x 40.00 grams NaOH / mole NaOH = 6.76 grams NaOH / L.
Assume the density of the solution is 1.00 since the solution is dilute. Therefore 1 liter of solution weighs 1,000 grams. Determine the percent of alkaline as follows: 6.76 g NaOH / 1000 g x 100 = 0.676 percent by weight.