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 pKa for TrisH+ is 8.07.
B. If 7.34 mL of 0.881 M NaOH is added, what is the pH after equilibration? (6 pts)
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
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