When 280 mL of 1.50 x 10^-4 M hydrochloric acid is added to 135 mL of 1.75 x 10^-4 M Mg(OH)2, the resulting solution will be? a) acidic b) basic c) neutral d) it is impossible to tell from the information given.

1 Answer
Apr 12, 2018

The resulting solution will be b) basic.

Explanation:

We have the volumes and concentrations of two reactants, so this is a limiting reactant problem.

Step 1. Gather all the information in one place

#color(white)(mmmmmmmmmm)"2HCl"color(white)(m) + color(white)(ll)"Mg(OH)"_2 → "MgCl"_2 + "2H"_2"O"#
#V"/mL":color(white)(mmmmmmll)280color(white)(mmmmll)135#
#"Conc./mol·L"^"-1":color(white)(m)1.50 ×10^"-4"color(white)(m)1.75×10^"-4"#

Step 2. Calculate the moles of each reactant

#"Moles of HCl" = 280 color(red)(cancel(color(black)("mL HCl"))) × (1.50 × 10^"-4"color(white)(l) "mmol HCl")/(1 color(red)(cancel(color(black)("mL HCl")))) = "0.042 00 mmol HCl"#

#"Moles of Mg(OH)"_2 = 135 color(red)(cancel(color(black)("mL Mg(OH)"_2))) × (1.75 × 10^"-4" color(white)(l)"mmol Mg(OH)"_2)/(1 color(red)(cancel(color(black)("mL Mg(OH)"_2)))) = "0.023 62 mmol Mg(OH)"_2#

Step 3. Identify the limiting reactant

Calculate the moles of #"MgCl"_2# we can obtain from each reactant.

From #"HCl"#:

#"Moles of MgCl"_2 = "0.042 00" color(red)(cancel(color(black)("mmol HCl"))) × "1 mmol MgCl"_2/(2 color(red)(cancel(color(black)("mmol HCl")))) = "0.021 00 mmol MgCl"_2#

From #"Mg(OH)"_2#:

#"Moles ofMgCaCl"_2 = "0.023 62"color(red)(cancel(color(black)("mmol Mg(OH)"_2))) × "1 mmol MgCl"_2/(1 color(red)(cancel(color(black)("mol Mg(OH)"_2)))) = "0.023 62 mmol MgCl"_2#

#"HCl"# is the limiting reactant because it gives the smaller amount of #"MgCl"_2#.

Step 4. Determine the pH of the solution

If #"HCl"# is the limiting reactant, #"Mg(OH)"_2# is the excess reactant.

#"Mg(OH)"_2# is a base, so the solution will be basic.