Respuesta :
The question is incomplete, the complete question is:
The solubility of slaked lime, [tex]Ca(OH)_2[/tex], in water is 0.185 g/100 ml. You will need to calculate the volume of [tex]2.50\times 10^{-3}[/tex]M HCl needed to neutralize 14.5 mL of a saturated
Answer: The volume of HCl required is 290mL, the mass of [tex]Ca(OH)_2[/tex] is 0.0268g, the moles of
Explanation:
Given values:
Solubility of [tex]Ca(OH)_2[/tex] = 0.185 g/100 mL
Volume of [tex]Ca(OH)_2[/tex] = 14.5 mL
Using unitary method:
In 100 mL, the mass of [tex]Ca(OH)_2[/tex] present is 0.185 g
So, in 14.5mL. the mass of [tex]Ca(OH)_2[/tex] present will be =[tex]\frac{0.185}{100}\times 14.5=0.0268g[/tex]
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)
Given mass of [tex]Ca(OH)_2[/tex] = 0.0268 g
Molar mass of [tex]Ca(OH)_2[/tex] = 74 g/mol
Plugging values in equation 1:
[tex]\text{Moles of }Ca(OH)_2=\frac{0.0268g}{74g/mol}=0.000362 mol[/tex]
Moles of [tex]OH^-[/tex] present = [tex](2\times 0.000362)=0.000724mol[/tex]
The chemical equation for the neutralization of calcium hydroxide and HCl follows:
[tex]Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O[/tex]
By the stoichiometry of the reaction:
Moles of [tex]OH^-[/tex] = Moles of [tex]H^+[/tex] = 0.000724 mol
The formula used to calculate molarity:
[tex]\text{Molarity of solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (mL)}}[/tex] .....(2)
Moles of HCl = 0.000724 mol
Molarity of HCl = [tex]2.50\times 10^{-3}[/tex]
Putting values in equation 2, we get:
[tex]2.50\times 10^{-3}mol=\frac{0.000724\times 1000}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{0.000725\times 1000}{2.50\times 10^{-3}}=290mL[/tex]
Hence, the volume of HCl required is 290mL, the mass of [tex]Ca(OH)_2[/tex] is 0.0268g, the moles of
