Question

Calculate the pH of the acetate buffer after 4 mL of 0.100 M HCl has been added.

Acetate buffer is made with 0.160 M NaC2H3O2 & 0.200 M HC2H3O2. We also took 10 mL of acetate buffer and diluted with 30mL of distilled water at the beginning.

Ka of HC2H3O2 is 1.8 x 10^-5.

I tried doing the SRF table and the pH I got was higher than the pH at the beginning so I know I am doing something wrong.

Answer 1

The approach to the problem will be to find the final concentration of acetic acid and sodium acetate, and then use Henderson-Hasselbalch equation to find the pH.

Now,

Concentration of NaC2H3O2 in the buffer = 0.160 M = 0.160 mol L^-1

Volume of buffer solution= 10 mL = 10*0.001 L =0.01 L ------------------[ 1 mL =0.001 L]

No. of moles of NaC2H3O2 in the buffer = 0.160 mol L^-1 * 0.01 L = 0.0016 moles

Concentration of  HC2H3O2 in the buffer = 0.200 M = 0.200 mol L^-1

Volume of buffer solution= 10 mL = 10*0.001 L =0.01 L   ------------------[ 1 mL =0.001 L]

No. of moles of HC2H3O2 in the buffer =0.200 mol L^-1 * 0.01 L = 0.002 moles

Concentration of HCl solution added =0.100 M = 0.100 mol L^-1

Volume of HCl solution added = 4 mL = 4 *0.001 L = 0.004 L   ------------------[ 1 mL =0.001 L]

No. of moles of HCl added = 0.100 mol L^-1 * 0.004 L = 0.0004 moles

The added HCl will react with NaC2H3O2 according to the following equation

NaC2H3O2 + HCl HC2H3O2 + NaCl

Therefore, there will be decrease in concentration of NaC2H3O2 and increase in concentration of HC2H3O2.

Constructing an ICE table to find no. of moles of sodium acetate and acetic acid at equilibrium

	NaC2H3O2	HCl	HC2H3O2
Initial no. of moles	0.00160 moles	0.0004 moles	0.0020 moles
Change in no. of moles	-0.0004 moles	-0.0004 moles	+0.0004 moles
Equilibrium no. of moles	0.0012 moles	0 moles	0.0024 moles

Total volume of the solution = 10 mL buffer + 30 mL water + 4 mL HCl

= 44 mL = 0.044 L ------------------[ 1 mL =0.001 L]

Equilibrium concentration of NaC2H3O2 =0.0012 moles / 0.044 L = 0.0273 mol L^-1

Equilibrium concentration of HC2H3O2 =0.0024 moles / 0.044 L = 0.0546 mol L^-1

Using Henderson Hasselbalch equation to find pH, we get

pH = pKa + log{ [NaC2H3O2] / [HC2H3O2]}

= - log ( 1.8* 10^-5) + log { (0.0273 / 0.0546)}

=4.745 +log 0.5

=4.745 -0.3010

=4.444