Question

A solution is prepared by mixing 150 mL of a 0.180 M tris solution with 200 mL of a 0.150 M Tris hydrochloride solution. The pK_a for TrisH⁺ is 8.07.

1. What is its pH (report to hundredth decimal place)? (5 pts)

B. If 7.34 mL of 0.881 M NaOH is added, what is the pH after equilibration? (6 pts)

Answer 1

Solution 1: 150 mL of a 0.180 M tris

Solution 2 : 200 mL of a 0.150 M Tris hydrochloride

After mixing solution 1 and solution 2 we get buffer solution. Volume of buffer solution will be 350 ml = 0.350 L

We can calculate [ tris ] and [ trisH+ ] in buffer solution by using dilution formula.

We have dilution formula, M stock V stock = M dilute V dilute

M dilute = M stock V stock / V dilute

Calculation for [ tris ]

M dilute = 0.180 M 150 ml / 350 ml = 0.0771 M

Calculation for [ trisH+ ]

M dilute = 0.150 M 200 ml / 350 ml = 0.0857 M

We have Henderson's equation, pH = pKa + log [ Base ] / [ Conjugate acid ]

pH = pKa + log [ Tris ] / [ Tris H+]

pH = 8.07 + log 0.0771 / 0.0857

pH = 8.07 - 0.04593

pH = 8.024

ANSWER : pH = 8.02

B)

Calculation of moles of tris, trisH+ and NaOH

We have relation, Molarity = No. of moles of solute / volume of solution in L

No. of moles of solute = Molarity   volume of solution in L

No. of moles of tris = 0.0771 mol / L 0.350 L = 0.026985 mol

No. of moles of trisH + = 0.0857 mol / L 0.350 L = 0.02995 mol

No. of moles of NaOH = 0.881 mol / L 0.00734 L = 0.00646654 mol

Consider reaction of NaOH with buffer solution.

trisH + + OH - tris + H2O

Let's use ICE table.

Moles	trisH + + OH - tris
I	0.02995	0.00646654	0.026985
C	- 0.00646654	- 0.00646654	+ 0.00646654
E	0.023483	0.00	0.033452

We have equation, pH = pKa + log [ Tris ] / [ Tris H+]

pH = 8.07 + log 0.033452 / 0.023483

pH = 8.07 + 0.15366

pH = 8.22

ANSWER : pH = 8.22