In the following acid-base neutralization, 1.61 g of the solid acid HC₆H₅O neutralized 11.61 mL of aqueous NaOH solution base by the reaction NaOH(aq) + HC₆H₅O(aq) $\rightarrow$ H₂O(l) + NaC₆H₅O(aq)
Calculate the molarity of the base solution.

Question

In the following acid-base neutralization, 1.61 g of the solid acid HC₆H₅O neutralized 11.61 mL of aqueous NaOH solution base by the reaction NaOH(aq) + HC₆H₅O(aq)

\rightarrow

H₂O(l) + NaC₆H₅O(aq)
Calculate the molarity of the base solution.

Quizplus · Accepted Answer

The correct answer is A because the molarity of the base solution can be calculated using the formula:Molarity = moles of solute / liters of solutionThe moles of solute can be calculated using the mass of the solute and its molar mass:Moles of solute = mass of solute / molar massThe molar mass of HC6H5O is 122.12 g/mol.The mass of solute is 1.61 g.Therefore, the moles of solute is:Moles of solute = 1.61 g / 122.12 g/mol = 0.0132 molThe volume of the solution is 11.61 mL, which is equal to 0.01161 L.Therefore, the molarity of the base solution is:Molarity = 0.0132 mol / 0.01161 L = 1.13 M

In the Following Acid-Base Neutralization, 1 →\rightarrow→ H2O(l) + NaC6H5O(aq) Calculate the Molarity of the Base Solution

In the Following Acid-Base Neutralization, 1 $\rightarrow$ H₂O(l) + NaC₆H₅O(aq)
Calculate the Molarity of the Base Solution