a) What volume of 10.0 M N a O H is needed to prepare a buffer with a p H of 7.7

Question

a) What volume of 10.0 M NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 g of TrisHCl?

Answer: 6.7 mL is needed to prepare buffer with pH of 7.79

b)The buffer from Part A is diluted to 1.00 L. To half of it (500. mL), you add 0.0150mol of hydrogen ions without changing the volume. What is the pH of the final solution?

Answer: pH of the final solution is 7.44

c) What additional volume of 10.0 M HCl would be needed to exhaust the remaining capacity of the buffer after the reaction described in Part B?

Please help with this one. Thank you!!

Explanation / Answer

When you react NaOH with TrisHCl, you will produce water and the original base, Tris. The presence of both Tris and TrisHCl in the same solution constitutes a buffer system (a weak base (Tris) plus its conjugate acid (TrisH+)). We need to calculate what the Tris / TrisH+ ratio is using the Henderson-Hasselbalch equation.

pH = pKa + log (moles Tris / moles TrisH+)

pKb for Tris = 5.92 (I looked it up); so pKa for TrisH+ = 14.00 - pKb = 14.00 - 5.92 = 8.08

pH = pKa + log (moles Tris / moles TrisH+)
7.79 = 8.08 + log (moles Tris / moles TrisH+)
-0.29 = log (moles Tris / moles TrisH+)
(moles Tris / moles TrisH+) = 10^-0.29 = 0.513

So, moles Tris = (0.513)(moles TrisH+)

The reaction between TrisH+Cl- and NaOH:

TrisH+Cl- + NaOH ==> Tris + NaCl . . .notice that all reactants and products are 1n a 1:1 mole ratio.

The molar mass for TrisHCl = 157.6 (I looked it up)

31.52 g TrisHCl x (1 mole TrisHCl / 157.6 g TrisHCl) = 0.2000 moles TrisHCl

So during the reaction, the TrisHCl (TrisH+) will be converted to Tris. But the sum of moles TrisHCl + moles Tris will always equal 0.2000.

moles Tris + moles TrisH+ = 0.2000

from above, (0.513)(moles TrisH+) = moles Tris . . .substitute that into the equation above.

(0.513)(moles TrisH+) + moles Tris = 0.2000
(1.513)(moles TrisH+) = 0.2000
moles TrisH+ = 0.2000 / 1.513 = 0.1322 moles TrisH+

moles Tris = 0.2000 - 0.1322 = 0.0678 moles Tris

So 0.0678 moles of Tris were produced in the reaction between TrisH+ and NaOH. That means that 0.0678 moles of NaOH were added since all reactants and products are in a 1:1 mole ratio.

moles NaOH = M NaOH x L NaOH
0.0678 = (10.0)(L NaOH)
L NaOH = 0.0678 / 10.0 = 0.00678 L = 6.78 mL

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