. 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
the answer is actually C since it is 0.030 L because Sr(OH)2 doubles it
The beaker that requires the smallest volume of 0.1 M HCl for complete neutralization is Beaker 4, which contains 15 mL of 0.1 M Sr(OH)2. This is because the moles of HCl required for neutralization are determined by the stoichiometry of the balanced equation.
For Sr(OH)2, the balanced equation is:
Sr(OH)2 + 2HCl -> SrCl2 + 2H2O
The stoichiometry shows that 1 mole of Sr(OH)2 reacts with 2 moles of HCl.
In Beaker 4, there is 0.015 L (15 mL) of 0.1 M Sr(OH)2, which is equal to:
0.015 L x 0.1 mol/L = 0.0015 mol Sr(OH)2
Since the stoichiometry is 1:2 (Sr(OH)2:HCl), the number of moles of HCl required for neutralization is:
0.0015 mol Sr(OH)2 x 2 mol HCl/1 mol Sr(OH)2 = 0.003 mol HCl
To convert this to volume, we divide by the concentration (0.1 M):
0.003 mol HCl / 0.1 mol/L = 0.03 L = 30 mL
Therefore, Beaker 4 requires the smallest volume of 0.1 M HCl for complete neutralization.