DO not undertsand part c), please help. What volume of 10.0 M NaOH is needed to

Question

DO not undertsand part c), please help.

What volume of 10.0 M NaOH is needed to prepere a buffer with a pH of 7.79 using 31.52 g of TrisHCl? Express your answer in milliliters using two significant figures. View Avaliable HInt(s) 6.7 mL Correct This procedure represents only one of many ways to prepare a buffer. The exact method chosen depends on the reagents and glassware available as well as the accuracy required for the pH. PartB The buffer from Part A is diluted to 1.00 T. To half of it (500. mT.) you add 0.0200 mol of hydrogen ions without changing the volume. What is the pH of the final solution? Express your answer numerically to two decimal places. View Available HInt(s) pll = 7.28 V Correct Notice that even though significant amount of the strong acid es added to the buffer the p did not change d as cally si ce the capacity o the bu er had ot been exceeded Part C What additional volume of 10.0 M HCl would be needed to exhaust the remaining capacity of the buttfer atter the reaction described in Part B? In other words, how much more of this HCI solution is required to consume the remaining Tris in the buffer? Express your answer in milliliters using two significant figures. View Available Hint(s) 1.8 mL Submit Previous Answers

Explanation / Answer

part a)

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

part c)

it's 2.34 mL.
Find your number of moles from your ice chart, then use C = n/V, deviated to V = n/C x 1000mL = 2.34 mL

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