Question

Determine the PH at the point in the titration of 40.0ml of 0.200M HC4H7o2 with 0.100 M Sr(OH)2 after 100ml of the strong base has been added. The value of Ka for HC4H7o2 is 1.5*10^-5

Answer 1

After adding 100 mL of 0.100 M Sr(OH)2 to 40 mL of 0.200 M HC4H7O2, the pH is determined to be 12.90, indicating the solution is basic due to the excess of OH- ions.

Step-by-step explanation:

The question involves determining the pH at a certain point during a titration process between a weak acid (HC4H7O2) and a strong base (Sr(OH)2). Given the initial concentration of the weak acid and the volume and concentration of the strong base added, along with the acid dissociation constant (Ka) of the weak acid, we can calculate the pH after the addition of the base.

First, determine the amount of moles of HC4H7O2 and Sr(OH)2:

• Moles of HC4H7O2 = 0.040 L × 0.200 M = 0.008 moles
• Moles of Sr(OH)2 added = 0.100 L × 0.100 M = 0.010 moles

Sr(OH)2 is a strong base that dissociates completely and since Sr(OH)2 provides two OH- ions per molecule, the total moles of OH- added would be 0.020 moles. This is greater than the moles of HC4H7O2, indicating that the equivalence point has been passed, and the solution is now basic with excess OH-.

To find the pH, we need to calculate the concentration of the excess OH- ions and then calculate the pOH, followed by the pH:

1. Excess moles of OH- = 0.020 - 0.008 = 0.012 moles
2. Concentration of OH- = 0.012 moles / (0.040 L + 0.100 L) = 0.08 M
3. pOH = -log(0.08) ≈ 1.10
4. pH = 14.00 - pOH ≈ 14.00 - 1.10 ≈ 12.90

The pH at the point in the titration after 100 mL of 0.100 M Sr(OH)2 has been added to 40.0 mL of 0.200 M HC4H7O2 is 12.90.

Answer 2

Check the explanation

Step-by-step 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).