Answer :
If a 25.0 ml sample of a saturated [tex]Ca(OH)_2[/tex] solution is titrated with 0.028 m HCl and the equivalence point is reached after 38.1 ml of titrant are dispensed, then the concentration of the saturated [tex]Ca(OH)_2[/tex] solution is approximately 0.021 M.
To find the concentration of the saturated [tex]Ca(OH)_2[/tex] solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between calcium hydroxide [tex]Ca(OH)_2[/tex] and hydrochloric acid (HCl):
[tex]Ca(OH)_{2} + 2HCl \rightarrow CaCl_{2} + 2H_{2}O[/tex]
Given:
Volume of [tex]Ca(OH)_2[/tex] solution (V_ca(OH)2) = 25.0 mL = 0.025 L
Volume of HCl titrant used (V_HCl) = 38.1 mL = 0.0381 L
Molarity of HCl titrant (M_HCl) = 0.028 M
At the equivalence point of the titration, the moles of HCl added will be equal to the moles of [tex]Ca(OH)_2[/tex] in the solution:
Moles of [tex]Ca(OH)_2[/tex] = Moles of HCl
Moles of HCl = Molarity of HCl × Volume of HCl titrant
Moles of [tex]Ca(OH)_2[/tex] = Moles of HCl / 2 (using the balanced equation)
Now, we can calculate the concentration of the [tex]Ca(OH)_2[/tex] solution:
Concentration of [tex]Ca(OH)_2[/tex] (M_ca(OH)2) = Moles of [tex]Ca(OH)_2[/tex] / Volume of Ca(OH)2 solution
M_ca(OH)2 = (Molarity of HCl × Volume of HCl titrant) / (2 × Volume of Ca(OH)2 solution)
M_ca(OH)2 = (0.028 M × 0.0381 L) / (2 × 0.025 L) ≈ 0.021 M.
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The actual question is:
A 25.0 ml sample of a saturated Ca(OH)₂ solution is titrated with 0.028 m HCl and the equivalence point is reached after 38.1 ml of titrant are dispensed. what is the concentration
