Final answer:
To neutralize 65.0 mL of 0.0400 M HCl, 68.4 mL of 0.0190 M Sr(OH)2 is required. This is calculated using stoichiometry, by first determining the number of moles of HCl and then finding the corresponding volume of Sr(OH)2 needed.
Step-by-step explanation:
The question involves a stoichiometry calculation within the subject of Chemistry to determine the volume of a strontium hydroxide, Sr(OH)2, solution needed to neutralize hydrochloric acid, HCl.
The balanced equation for the reaction is:
- Sr(OH)2 + 2HCl → SrCl2 + 2H2O
From the stoichiometry of the reaction, we see that one mole of Sr(OH)2 neutralizes two moles of HCl.
To find the volume of Sr(OH)2 needed, we use the following steps:
- Calculate the number of moles of HCl: mol HCl = 0.0400 M * 0.065 L = 0.0026 mol
- Use the stoichiometry to find the moles of Sr(OH)2: mol Sr(OH)2 = 0.5 * mol HCl = 0.0013 mol
- Calculate the volume of Sr(OH)2: Volume = mol Sr(OH)2 / Molarity Sr(OH)2 = 0.0013 mol / 0.0190 M = 0.0684 L = 68.4 mL
Thus, 68.4 mL of 0.0190 M Sr(OH)2 would be required to neutralize 65.0 mL of 0.0400 M HCl.