Q94E
Question
Potassium acid phthalate, KHC6H4O4 or KHP, is used in many laboratories, including general chemistry laboratories, to standardize solutions of base. KHP is one of only a few stable solid acids that can be dried by warming and weighed. A 0.3420-g sample of KHC6H4O4 reacts with 35.73 mL of a NaOH solution in a titration. What is the molar concentration of the NaOH?
\({\rm{KH}}{{\rm{C}}_6}{{\rm{H}}_4}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + NaO}}{{\rm{H}}_{({\rm{aq}})}}{\rm{ }} \to {\rm{ KNa}}{{\rm{C}}_6}{{\rm{H}}_4}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + }}{{\rm{H}}_2}{{\rm{O}}_{({\rm{aq}})}}\)
Step-by-Step Solution
VerifiedThe molar concentration of sodium hydroxide is \({\rm{5}}{\rm{.3 }} \times {\rm{ 1}}{{\rm{0}}^{ - 3}}\;{\rm{M}}\)
According to the question, it is given that
The mass of KHC6H4O4 is 0.3420 g
The molecular mass of KHC6H4O4 is \(39\,{\rm{ + }}{\rm{5 }} \times {\rm{ 1 + }}{\rm{12 }} \times {\rm{ 6 + }}{\rm{16 }} \times {\rm{ }}4\;{\rm{ = 180 g/mol}}\)
Thus, the number of moles of KHC6H4O4 will be given by:
\(\begin{array}{l}{\rm{Moles = }}\frac{{{\rm{Mass}}}}{{{\rm{Molecular mass}}}}\\{\rm{Moles = }}\frac{{0.3420}}{{{\rm{180}}}}{\rm{ = 0}}{\rm{.0019 moles}}\end{array}\)
Now, from the reaction we can determine that when 1 mole of KHC6H4O4 reacts with sodium hydroxide it requires 1 mole of NaOH, thus the number of moles of sodium hydroxide will be:
\({\rm{1 }} \times {\rm{ 0}}{\rm{.0019 = 0}}{\rm{.0019 moles}}\)
The molar concentration of a compound, also known as molarity, is a method of determining the concentration of a substance in a solvent. As the number of moles of sodium hydroxide is 0.0019 moles and the volume is 35.73 ml. Thus, the molar concentration of sodium hydroxide is given by:
\(\begin{array}{l}{\rm{Concentration = }}\frac{{{\rm{Moles}}}}{{{\rm{Volume}}}}\\{\rm{Concentration = }}\frac{{0.0019}}{{35.73}}{\rm{ = 5}}{\rm{.3 }} \times {\rm{ 1}}{{\rm{0}}^{ - 3}}\;{\rm{M}}\end{array}\)