Final answer:
To find the molarity of the original sodium hydroxide solution, calculate the moles of HCl initially added, determine the moles of HCl remaining after the first reaction, subtract to find the moles of HCl that reacted with NaOH, and finally divide by the volume of NaOH to find its molarity. The calculated molarity is closest to 0.91 M, option D.
Step-by-step explanation:
To determine the molarity of the original sodium hydroxide solution, we must first understand the reaction equations involved in the titrations:
As the question states, the resulting solution after the first titration is acidic, indicating that there wasn't enough NaOH to completely neutralize the HCl. This means some HCl remained to be neutralized by the subsequent addition of barium hydroxide, where the reaction ratio of HCl to Ba(OH)2 is 2:1.
First, we need to determine the moles of HCl initially added:
33.7 mL of 1.54 M HCl contains (33.7 mL * 1.54 M / 1000) = 0.051898 moles of HCl.
Since 16.4 mL of 1.19 M Ba(OH)2 were required to neutralize the remaining HCl, we can calculate the moles of HCl that remained after the reaction with NaOH:
(16.4 mL * 1.19 M / 1000) = 0.019516 moles of Ba(OH)2, which neutralizes twice the amount of HCl (because of the 2:1 ratio), resulting in 0.039032 moles of HCl remaining.
Therefore, the moles of HCl that reacted with NaOH initially are 0.051898 - 0.039032 = 0.012866 moles.
Knowing the moles of HCl that reacted with NaOH and the volume of NaOH used, we can calculate the molarity of NaOH:
Molarity of NaOH = Moles of HCl reacted / Volume of NaOH in L = 0.012866 moles / 0.0312 L = 0.4120513 M, which suggests the closest answer is (D) 0.91 M.