Question

Under the same conditions how many mL of ${\text{1 M KOH}}$ and ${\text{0}}{\text{.5 M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ solutions, respectively, when mixed to form a total volume of 100 mL produce the highest rise in temperature?
(A) 67, 33
(B) 33, 67
(C) 40, 60
(D) 50, 50

Answer 1

For each option, calculate the number of milli-moles of ${\text{KOH}}$ that can be neutralized. Select the option in which the maximum number of milli-moles of ${\text{KOH}}$ that can be neutralized.

Complete step by step answer:
In the question, you are given the molar concentrations of potassium hydroxide and sulphuric acid solutions. You are given a total volume of two solutions. You are also given four options in which different values of volumes of two solutions are given. You are asked to determine the volumes of two solutions that should be added in order to get maximum temperature rise.
When you multiply the molar concentration of a solution with volume in milliliter, you get the number of millimoles.
${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ is sulphuric acid. It is a dibasic acid. It has two ionizable protons. ${\text{KOH}}$ is potassium hydroxide. It is a monoacidic base. It has only one ionizable hydroxide ion.
The reaction of sulphuric acid with potassium hydroxide is acid base neutralization reaction.
Write the balanced chemical equation for the reaction between sulphuric acid and potassium hydroxide.
${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} + 2{\text{KOH}} \to {{\text{K}}_2}{\text{S}}{{\text{O}}_4}{\text{ + 2}}{{\text{H}}_2}{\text{O}}$
Thus, one molecule of sulphuric acid reacts with two molecules of potassium hydroxide to form one molecule of potassium sulphate and two molecules of water.
(A) Suppose you react ${\text{67mL}}$ of ${\text{1 M KOH}}$ and ${\text{33mL}}$ of ${\text{0}}{\text{.5 M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ solutions
Calculate the number of millimoles of ${\text{KOH}}$
${\text{67mL }} \times {\text{ 1 mmol/mL = 67mmol}}$
Calculate the number of millimoles of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$
${\text{33mL }} \times {\text{ 0}}{\text{.5 mmol/mL = 16}}{\text{.5mmol}}$
${\text{16}}{\text{.5mmol}}$ of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ will neutralize ${\text{16}}{\text{.5mmol}} \times 2 = {\text{33mmol}}$ of ${\text{KOH}}$ .
Hence, the maximum number of milli-moles of ${\text{KOH}}$ that can be neutralized is ${\text{33mmol}}$ .

(B) Suppose you react ${\text{33mL}}$ of ${\text{1 M KOH}}$ and ${\text{67mL}}$ of ${\text{0}}{\text{.5 M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ solutions
Calculate the number of millimoles of ${\text{KOH}}$
${\text{33mL }} \times {\text{ 1 mmol/mL = 33mmol}}$
Calculate the number of millimoles of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$
${\text{67mL }} \times {\text{ 0}}{\text{.5 mmol/mL = 33}}{\text{.5mmol}}$
${\text{33}}{\text{.5mmol}}$ of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ will neutralize ${\text{33}}{\text{.5mmol}} \times 2 = 67{\text{mmol}}$ of ${\text{KOH}}$ .
But only ${\text{33mmol}}$ of ${\text{KOH}}$ are present.
Hence, the maximum number of milli-moles of ${\text{KOH}}$ that can be neutralized is ${\text{33mmol}}$ .
(C) Suppose you react ${\text{40mL}}$ of ${\text{1 M KOH}}$ and ${\text{60mL}}$ of ${\text{0}}{\text{.5 M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ solutions
Calculate the number of millimoles of ${\text{KOH}}$
${\text{40mL }} \times {\text{ 1 mmol/mL = 40mmol}}$
Calculate the number of millimoles of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$
${\text{60mL }} \times {\text{ 0}}{\text{.5 mmol/mL = 30mmol}}$
${\text{30mmol}}$ of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ will neutralize ${\text{30mmol}} \times 2 = 60{\text{mmol}}$ of ${\text{KOH}}$ .
But only ${\text{40mmol}}$ of ${\text{KOH}}$ are present.
Hence, the maximum number of milli-moles of ${\text{KOH}}$ that can be neutralized is ${\text{40mmol}}$ .
(D) Suppose you react ${\text{50mL}}$ of ${\text{1 M KOH}}$ and ${\text{50mL}}$ of ${\text{0}}{\text{.5 M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ solutions
Calculate the number of millimoles of ${\text{KOH}}$
${\text{50mL }} \times {\text{ 1 mmol/mL = 50mmol}}$
Calculate the number of millimoles of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$
${\text{50mL }} \times {\text{ 0}}{\text{.5 mmol/mL = 25mmol}}$
${\text{25mmol}}$ of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ will neutralize ${\text{25mmol}} \times 2 = 50{\text{mmol}}$ of ${\text{KOH}}$ .
Hence, the maximum number of moles of ${\text{KOH}}$ that can be neutralized is ${\text{50mmol}}$ .
Hence, in the option (D), the maximum number of milli-moles of ${\text{KOH}}$ that can be neutralized.

Hence, the correct answer is the option (D) 50, 50.

Note: The acid base neutralization reaction is an exothermic process. Heat energy is given out during the process. For a fixed total volume of the mixture of sulphuric acid and potassium hydroxide, you can obtain the highest temperature rise for a mixture in which the maximum number of moles of ${\text{KOH}}$ are neutralized.