Answer:
Check the explanation
Explanation:
Mols HC4H7O2 = (volume in L)*(molarity) = (40.0 mL)*(0.200 M)
= (40.0 mL)*(1 L)/(1000 mL)*(0.200 M)
= 8.00*10-3 mol.
Mols Sr(OH)2 corresponding to 10.0 mL of 0.100 M solution =
(volume in L)*(molarity)
= (10.0 mL)*(0.100 M)
= (10.0 mL)*(1 L)/(1000 mL)*(0.100 M)
= 1.00*10-3 mol.
Consider the ionization of Sr(OH)2 as below.
Sr(OH)2 (aq) ----------> Sr2+ (aq) + 2 OH- (aq)
As per the stoichiometric equation,
1 mol Sr(OH)2 = 2 mols OH-.
Therefore,
0.0010 mol Sr(OH)2 = [0.0010 mol Sr(OH)2]*(2 mols OH-)/[1 mole Sr(OH)2]
= 0.0020 mol
= 2.00*10-3 mol
Set up the ICE charts as below.
HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)
Before (mol) 8.00*10-3 2.00*10-3 - -
Change (mol) -2.00*10-3 -2.00*10-3 - +2.00*10-3
After (mol) 6.00*10-3 0 - 2.00*10-3
The change in a pure substance, e.g., H2O is not considered in an acid-base reaction.
Volume of the solution = (40.0 + 10.0) mL = 50.0 mL = (50.0 mL)*(1 L)/(1000 mL) = 0.05 L.
The initial concentrations are obtained by dividing the numbers of moles by the volume, 0.05 L.
Set up the ICE charts as below.
HC4H7O2 (aq) + OH- (aq) ------------> H2O (l) + C4H7O2- (aq)
Initial (M) 0.160 0.0400 - -
Change (M) -0.0400 -0.0400 - +0.0400
Equilibrium (M) 0.120 0 - 0.0400
The acid-ionization constant is written as
Ka = [H3O+][C4H7O2-]/[HC4H7O2] = 1.5*10-5
Plug in the known values and get
Ka = [H3O+]*(0.0400)/(0.120) = 1.5*10-5
======> [H3O+] = (1.5*10-5)*(0.120)/(0.0400) (ignore units)
======> [H3O+] = 4.5*10-5
The proton concentration of the solution is 4.5*10-5 M.
pH = -log (4.5*10-5 M)
= 4.346
≈ 4.35 (ans).
