280. mL of 1.50 × 10-2 mol L-1 hydrochloric acid is added to 135 mL of 1.75 × 10-2 mol L-1 Mg(OH)2. Calculate the volume of 5.25 x 10-5 mol L-1 of HCl to be added to this solution to make it a neutral solution.
To solve this problem, we can use the concept of stoichiometry and the equation for the neutralization reaction between HCl and Mg(OH)2. The balanced chemical equation for this reaction is:
2HCl + Mg(OH)2 -> MgCl2 + 2H2O
According to the equation, 2 moles of HCl react with 1 mole of Mg(OH)2 to form 1 mole of MgCl2 and 2 moles of H2O.
First, let's calculate the number of moles of HCl and Mg(OH)2 in the initial solution using the given concentrations and volumes:
Moles of HCl = Concentration of HCl x Volume of HCl
= (1.50 × 10^-2 mol L^-1) x (280 mL) x (1 L / 1000 mL)
= 0.0042 mol
Moles of Mg(OH)2 = Concentration of Mg(OH)2 x Volume of Mg(OH)2
= (1.75 × 10^-2 mol L^-1) x (135 mL) x (1 L / 1000 mL)
= 0.00236 mol
Now, let's calculate the moles of HCl needed to neutralize the Mg(OH)2:
Moles of HCl needed = 2 x Moles of Mg(OH)2
= 2 x 0.00236 mol
= 0.00472 mol
To find the volume of 5.25 x 10^-5 mol L^-1 HCl solution needed to neutralize the initial solution, we can use the formula:
Volume of HCl needed = Moles of HCl needed / Concentration of HCl
Volume of HCl needed = 0.00472 mol / (5.25 x 10^-5 mol L^-1)
= 90.1 L
Therefore, the volume of 5.25 x 10^-5 mol L^-1 HCl needed to make the solution neutral is 90.1 L.