. Consider the following data table Beaker Volume Contents 50 mL 0.2 M NaOH 25 mL 0.1 M KOH 3 20 mL 0.2 M NH4OH 15 ml 0.1 M Sr(OH)2 Identify the beaker that requires the smallest volume of 0.1 M HCL for complete neutralization. A. Beaker 1 B. Beaker 2 C. Beaker 3 D. Beaker 4
To determine which beaker requires the smallest volume of 0.1 M HCl for complete neutralization, we need to calculate the number of moles of hydroxide ions in each beaker.
Beaker 1 contains 50 mL of a 0.2 M NaOH solution. The number of moles of NaOH in the beaker is:
0.2 moles/L * (50 mL / 1000 mL/L) = 0.01 moles
Beaker 2 contains 25 mL of a 0.1 M KOH solution. The number of moles of KOH in the beaker is:
0.1 moles/L * (25 mL / 1000 mL/L) = 0.0025 moles
Beaker 3 contains 20 mL of a 0.2 M NH4OH solution. The number of moles of NH4OH in the beaker is:
0.2 moles/L * (20 mL / 1000 mL/L) = 0.004 moles
Beaker 4 contains 15 mL of a 0.1 M Sr(OH)2 solution. The number of moles of Sr(OH)2 in the beaker is:
0.1 moles/L * (15 mL / 1000 mL/L) = 0.0015 moles
From these calculations, we can see that Beaker 1 contains the highest number of moles of hydroxide ions (0.01 moles). Therefore, Beaker 1 will require the smallest volume of 0.1 M HCl for complete neutralization.
So, the answer is A. Beaker 1.