How do I do the box on page 2? REPORT SHEET Buffers Section_ Name Nicole Rigby D

ID: 704655 • Letter: H

Question

How do I do the box on page 2? REPORT SHEET Buffers Section_ Name Nicole Rigby Date Iestractor Partner's Name Eauations (2 pts. each) Show the equation for the reaction of an acetate buffer with a strong acid represented by Ho Show the equation for the reaction of an acetate buffer with a strong base represented by OH (2 pts. each) .O713 Mass of sodium acetate in acetate buffer 3 Volume of 3.0 M acetic acid in acetate buffer 1. 2620 Mass of ammonium chloride 5.36' Volume of 5.0 M NH&OH; pHH Measurements. Experimental Values (2 pts. each) Ammonia Buffer Deionized Water Acetate Buffer pHf of solution6.7H.66 4.27 pH of solution after addition of 1 mL of 0.6 M NaOH 11209 | ?.34 | 9.69 pH of solution after addition of I mL of 0.6 M HCI | | ??? ?. 5?

Explanation / Answer

Let us start with the equations, since you seem to have got them a bit incorrect. Acetate buffer is composed of acetic acid (CH3COOH) and acetate salt (CH3COO-), usually as the sodium salt (sodium acetate, CH3COO-Na+). Sodium ion (Na+) acts as a spectator ion and doesn’t participate in the acid/base equilibrium. Hence, the active components of the buffer are CH3COOH and CH3COO-.

H3O+ is a strong acid and reacts with the base (acetate, CH3COO-) to give a weak acid (acetic acid, CH3COOH) and water (H2O) as per the reaction below.

CH3COO- (aq) + H3O+ (aq) --------> CH3COOH (aq) + H2O (l) ……(1)

OH- is a strong base and reacts with the acid (acetic acid, CH3COOH) to give a weak base (acetate, CH3COO-) and H2O as per the reaction below.

CH3COOH (aq) + OH- (aq) --------> CH3COO- (aq) + H2O (l) …..(2)

Molar mass of sodium acetate = 82.043 g/mol.

Mass of sodium acetate taken = 1.0713 g.

Moles of sodium acetate taken = (mass of sodium acetate)/(molar mass of sodium acetate) = (1.0713 g)/(82.043 g/mol) = 0.01308 mole ? 0.0131 mole.

Molarity of sodium acetate = (moles of sodium acetate)/(volume of buffer solution in L) = (0.0131 mole)/(0.250 L) = 0.0524 mol/L = 0.0524 M (ans).

Moles of acetic acid = (volume of acetic acid in L)*(molarity of acetic acid) = (3.62 mL)*(1 L/1000 mL)*(3.0 M) = 0.01086 mole ? 0.0110 mole

Molarity of acetic acid = (moles of acetic acid)/(volume of buffer solution in L) = (0.0110 mole)/(0.250 L) = 0.0440 mol/L = 0.0440 M (ans).

Finally look at the table. We add 1.00 mL of 0.6 M NaOH. Therefore,

moles of NaOH = (volume of NaOH in L)*(molarity of NaOH) = (1.00 mL)*(1 L/1000 mL)*(0.6 M) = 6.0*10-4 mole = 0.0006 mole.

NaOH ionizes in aqueous solution to give OH-. OH- reacts with CH3COOH as per equation (2) above.

As per the stoichiometric equation,

1 mole CH3COOH = 1 mole OH- = 1 mole CH3COO-.

Therefore,

0.0006 mole OH- = 0.0006 mole CH3COOH consumed = 0.0006 mole CH3COO- formed.

Now, fill up the ICE chart.

Moles

CH3COOH

OH-

CH3COO-

Before addition

0.0110

0.0006

0.0131

Addition

-0.0006 (0.0006 moles of CH3COOH is neutralized by OH-)

-0.0006 (OH- neutralizes CH3COOH)

0.0006 (0.0006 mole CH3COO- is formed by the action of OH-)

After addition

(0.0110 – 0.0006) = 0.0104

(0.0131 + 0.0006) = 0.0137