Calculate the pH of the solution made by mixing 35.0 mL of 0.500 M ca(OH)2 with 45.0 mL of 0.710 MHCl
Ca(OH)2 + 2HCl ==> CaCl2 + 2H2O
mols Ca(OH)2 = M x L = 0.500M x 0.035L = 0.0175 mols.
mols HCl = M x L = 0.710M x 0.045L = 0.03195
Since the reaction is 1 mol Ca(OH)2 to 2 mols HCl, that means if all the Ca(OH)2 reacts, it will require 0.0175 x (2 mol HCl/1 mol Ca(OH)2) = 0.0175 x 2/1 = 0.035 mols HCl and we don't have that much; therefore, the HCl is the limiting reagent, all of it will react, and some of the Ca(OH)2 will remain behind unreacted. How much will that be.
We do the reverse calculation for HCl.
0.03195 mols HCl x (1 mol Ca(OH)2/2 mols HCl) = 0.01598 mols Ca(OH)2 required. So how much is left.
0.01750-0.01598 = 0.00153 mols Ca(OH)2.
What is the volume will we have.
35.0 mL + 45.0 mL = 80.0 mL = 0.080 L so
[Ca(OH)2] = 0.00153/0.080 = 0.01906 M. Since there are two OH^- per Ca(OH)2 molecule, (OH^-) is twice that or 0.01906 x 2 = ??
pOH form that.
pH from that.
Post your work if you get stuck.
Check my arithmetic. These numbers all look alike on my keyboard.
To calculate the pH of the solution, we need to find the concentrations of the hydronium ions ([H3O+]) or hydroxide ions ([OH-]) in the solution.
First, let's determine the number of moles of HCl and Ca(OH)2:
Moles of HCl:
Molarity (M) = Moles (mol) / Volume (L)
0.710 M = Moles / 0.045 L
Moles of HCl = 0.710 M * 0.045 L = 0.03195 mol
Moles of Ca(OH)2:
Molarity (M) = Moles (mol) / Volume (L)
0.500 M = Moles / 0.035 L
Moles of Ca(OH)2 = 0.500 M * 0.035 L = 0.0175 mol
Now, we need to determine which reactant is the limiting reactant and calculate the remaining amount of OH- or H3O+ ions.
Since the balanced chemical equation between HCl and Ca(OH)2 is:
2HCl + Ca(OH)2 -> CaCl2 + 2H2O
The stoichiometric ratio is 2:1 for HCl to Ca(OH)2.
Comparing the moles of HCl and Ca(OH)2, we see that we have excess HCl since 0.03195 mol HCl > 2 * 0.0175 mol Ca(OH)2.
Now let's calculate the remaining moles of Ca(OH)2:
Remaining moles of Ca(OH)2 = Moles of Ca(OH)2 - (2 * Moles of HCl)
Remaining moles of Ca(OH)2 = 0.0175 mol - (2 * 0.03195 mol) = -0.0454 mol
Since we cannot have negative moles, the reaction consumes all the moles of Ca(OH)2, leaving no hydroxide ions ([OH-]) in solution since Ca(OH)2 dissociates into Ca^2+ and 2OH- ions.
Now, let's calculate the concentration of OH- ions:
Volume of the final solution = volume of HCl + volume of Ca(OH)2
Volume of the final solution = 0.045 L + 0.035 L = 0.08 L
Concentration of OH- ions = Moles of OH- ions / Volume of the final solution
Concentration of OH- ions = 0 mol / 0.08 L = 0 M
Since the hydroxide ion concentration is zero, and pH = -log10([H3O+]), the pH is undefined.