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Potassium acid phthalate, KHC6H4O4or 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?

\({\rm{KH}}{{\rm{C}}_6}{{\rm{H}}_4}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + NaO}}{{\rm{H}}_{({\rm{aq}})}}{\rm{ }} \to {\rm{ KNa}}{{\rm{C}}_6}{{\rm{H}}_4}{{\rm{O}}_{4({\rm{aq}})}}{\rm{ + }}{{\rm{H}}_2}{{\rm{O}}_{({\rm{aq}})}}\)

Short Answer

Expert verified

The molar concentration of sodium hydroxide is \({\rm{5}}{\rm{.3 }} \times {\rm{ 1}}{{\rm{0}}^{ - 3}}\;{\rm{M}}\)

Step by step solution

01

Moles of sodium hydroxide

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:

\({\rm{1 }} \times {\rm{ 0}}{\rm{.0019 = 0}}{\rm{.0019 moles}}\)

02

Molar concentration of sodium hydroxide

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:

\(\begin{array}{l}{\rm{Concentration = }}\frac{{{\rm{Moles}}}}{{{\rm{Volume}}}}\\{\rm{Concentration = }}\frac{{0.0019}}{{35.73}}{\rm{ = 5}}{\rm{.3 }} \times {\rm{ 1}}{{\rm{0}}^{ - 3}}\;{\rm{M}}\end{array}\)

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Most popular questions from this chapter

What is the concentration of NaCl in a solution if titration of 15.00 mL of the solution with 0.2503MAgNO3 requires 20.22 mL of the AgNO3 solution to reach the end point?

AgNO3(aq) + NaCl(aq)⟶A²µ°ä±ô(s) + NaNO3(aq)

Use the following equations to answer the next four questions:

i.\({H_2}O\left( s \right) \to {H_2}O\left( l \right)\)

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iii.\(C{H_3}OH\left( g \right) + {O_2}\left( g \right) \to C{O_2}\left( g \right) + {H_2}O\left( g \right)\)

iv. \(2{H_2}O\left( l \right) \to 2{H_2}\left( g \right) + {O_2}\left( g \right)\)

v. \({H^ + }\left( {aq} \right) + O{H^ - }\left( {aq} \right) \to {H_2}O\left( l \right)\)

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