Question

a) What is the pH of a buffer solution prepared by mixing 120 mL of a 0.435 M CH₃CO₂H with 50 mL of a .286 M CH₃CO₂Na solution? CH₃CO₂H K_a = 1.75 x 10^-5

b) What is the resulting pH if we add 10.2 mL of a .513 M HCl solution to this buffer? Does the pH make sense? Why/why not?

c) What is the resulting pH if we add 3.56 mL of a 0.420 M NaOH solution to the original buffer?

Answer 1

K_a = 1.75 x 10^-5

pKa = - log ka

= - log 1.75 x 10^-5

= 4.76

Moles of acid = molarity * volume in L

= 0.435 * 0.120

= 0.0522 moles

Moles of base = molarity * volume in L

= 0.286 * 0.050

= 0.0143 moles

pH of this buffer:

pH = pKa + log (base/acid)

pH = 4.76 + log (0.0143/ 0.0522)

pH = 4.76 +(-0.56)

= 4.18

b) What is the resulting pH if we add 10.2 mL of a .513 M HCl solution to this buffer? Does the pH make sense? Why/why not?

Moles of acid = 0.513 *0.0102

= 0.00523 moles HCl

Now calculate the new concentration of acid and base as follows:

base = 0.0143 -0.00523 / total volume, 0.1802 L

= 0.0503 M

acid = 0.0522 + 0.00523   / 0.1802 L

= 0.319 M

pH of this buffer:

pH = pKa + log (base/acid)

pH = 4.76 + log (0.00503/ 0.319)

pH = 4.76 +(-0.80)

= 3.94

c) What is the resulting pH if we add 3.56 mL of a 0.420 M NaOH solution to the original buffer?

Now calculate the mole of NaOH

= 0.420 mol / 1000 mL *3.56 Ml

=0.0015 mol

NaOH < => Na+ OH-

Now calculate the new concentration of acid and base as follows:

base = 0.0143 + 0.0015 / total volume, 0.174 L

= 0.09080 M

acid = 0.0522- 0.0015 / total volume, 0.174 L

= 0.291 M

pH = pKa + log (base/acid)

pH = 4.76 + log (0.09080/ 0.291)

pH = 4.76 +(-0.51)

= 4.25