Question

1a.) m‐Dimethylaminobenzoic acid (abbreviated mDAB‐H) is a weak acid with a pKa of 5.10. mDAB‐H has a molecular formula of C9H11NO2. If 22.333 g of mDAB‐H is dissolved in 325.0 mL of water, what is the pH and pOH of the resulting solution?

b.)A buffer solution using mDAB‐H and its sodium salt is to be made with a pH of 5.00. If 22.333 g of mDAB‐H is dissolved in 325.0 mL of water (the same as before), what mass of the sodium salt (mDAB‐Na) must be added to generate the desired buffer?

c.)To the buffer prepared in part b is added 50.0 mL of a 0.250 M NaOH solution. What is the resulting pH of the buffer?

d.) The buffer capacity is reached when the pH of the buffer shifts by 1.00 pH unit. What volume of a 1.00 M HCl solution would need to be added to a (fresh, from #4) mDAB‐H/mDAB‐Na buffer solution to reach this point?

Answer 1

Henderson–Hasselbalch equation for weak acid/base is given by

pH= pKa + log [A-]/[HA], [HA] is the molarity of undissociated weak acid, [A⁻] is the molarity of this acid's conjugate base.

it can also represented in terms of pOH, pOH = pKb + log [HA]/[A-], pH + pOH = 14.

a.

pKa of m‐Dimethylaminobenzoic acid (mDAB‐H) = 5.10

Molar mass of mDAB‐H = 165.192 g/mol

Mass. of mDAB‐H = 22.333 g

Volume of water = 325.0 mL = 0.325 L

No. of moles of mDAB‐H = 22.333 g/165.192 g/mol = 0.1352 mol

Molarity of mDAB‐H = 0.1352 mol / 0.325 L = 0.416 M

pKa = -log Ka

Ka = 10^-ka = 10^-5.10 = 7.943 x 10^-6

If x mole of mDAB‐H is ionizes to mDAB and H+, then

[mDABH] = (0.416 - x) mol

[mDAB] = x mol

[H3O+] = x mol

Ka = [mDAB][H3O+]/[mDAB-H] = x^2/(0.416 - x) = 7.943 x 10^-6

solving x, x = 0.001814

[H3O+] = x = 0.001814

pH = -log [H3O+] = -log (0.001814) = 2.74

pOH = 14-pH = 14 - 2.74 = 11.26

b.

pH of the buffer required = 5.00

pH= pKa + log [A-]/[HA]

5.00 = 5.10 + log [A-]/[HA]

log [A-]/[HA] = 5.00 - 5.10 = -0.1

[A-]/[HA] = 10^-0.1 = 0.7943

[HA] = 0.416 M (calculated above)

[A-] = 0.7943 x [HA] = 0.7943 x 0.416 = 0.3304 M

Molarity of mDAB‐Na required = 0.3304 M

Total volume = 325.0 mL

No. of moles of mDAB‐Na = 0.3304 mol/L x 0.325 L = 0.1074 mol

Molar mass of mDAB‐Na = 187.1710 g/mol

Mass of 0.1074 mol of mDAB‐Na = 187.1710 g/mol x 0.1074 mol = 20.1022 g

Mass of mDAB‐Na salt need to add = 20.1022 g

c.

No. of moles of NaOH in 50.0 mL of a 0.250 M = 0.250 M x 50.0 L / 1000 = 0.0125 mol

No. of moles of mDAB‐H = 0.1352 mol

On adding 0.0125 mol of NaOH (strong base), 0.0125 mol of mDAB‐H is converted to 0.0125 mol mDAB‐Na

No. of moles of mDAB‐H (as above) = 0.1352 mol

No. of moles of mDAB‐Na (as above) = 0.1074 mol

[mDAB‐H] = 0.1352 - 0.0125 = 0.1227 mol

[mDAB‐Na] = 0.1074 mol + 0.0125 mol = 0.1199 mol

pH = 5.10 + log [A-]/[HA] = 5.10 + log (0.1199/ 0.1227) = 5.10 - 0.01 = 5.09

pH after adding NaOH = 5.09

d.

Buffer pH of fresh buffer as in b = 5.00

pH change required = 1 unit

required pH = 5.00 - 1.00 = 4.00

4.00 = 5.10 + log [A-]/[HA]

log [A-]/[HA] = 4.00 - 5.10 = -1.10

[A-]/[HA] = 10^-1.10 = 0.07943

[A-]+[HA] = 0.1352 mol + 0.1074 mol = 0.2426

Solving,

[A-] = 0.01785 mol

[HA] = 0.22475 mol

Which means that 0.22475 mol - 0.1352 mol = 0.08955 mol of H+ ions need to be added

Molarity of HCl = 1.00 M

Volume of 1.00 M HCl equal to 0.08955 HCl = 0.08955 mol / 1.00 mol/L = 0.08955 L = 89.55 mL

Volume of a 1.00 M HCl solution need to add = 89.55 mL