Question

Part A and Part B are correct, need Part C. What additional volume of 10.0 M HCl would be needed to consume all the remaining base component of the buffer (Tris) after the reaction described in Part B?

Review Constants 1 Periodic Table In the study of biochemical processes, a common buffering agent is the weak base trishydroxymethylaminomethane, (HOCH),CNH, che abbreviated as Tris. At 25 C. Tris has a pks of 5.91. The hydrochloride of Tris is (HOCH), CNH,C1, which can be abbreviated as TrisHCI. What volume of 10.0 M NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 of TrisHCI? Express your answer in milliliters using two significant figures. View Available Hint(s) 6.7 ml Previous Answers 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 Part 8 The buffer from Part Als diluted to 1.00 L. To half of (500 ml), you add 0.0150 mol of hydrogen bons without changing the volume. What is the pll of the final solution? Express your answer numerically to two decimal places View Available Hint) pll - 7.44

Answer 1

When [H⁺] = [OH^-], the pH must be equal to 7.0 but the pH of remaining solution of part B is 7.44. I must calculate the delta pH to calculate the amount of H⁺ I have to add to the solution.

$\Delta pH:$ is the difference among actual pH and desire pH

$\Delta pH=7.44-7.00=0.44$

The $\Delta pH$ is the [H⁺] concentration variation   $\Delta pH=-\log \Delta [H^{+}]$

$\Delta [H^{+}]=10^{-\Delta pH}$

$\Delta [H^{+}]=10^{-0.44}=0.36\frac{mol H^{+}}{L_{added}}$

In part B solution there are 500ml whose [H⁺] concentration lacks in 0.36M to achieve the neutrality and I have to add:

$\tfrac{0.36molH^{+}}{L}*0.5L=nH^{+}=0.18mol$

Finally I have to apply mass conservation to calculate the volume I have to take from 10.0M HCl solutionto get these amount of n_HCl.

$[HCl]_{0}V_{0}=n_{HCl}$

The initial [HCl]₀ is 10.0M

V₀ I must calculate

$10.0\frac{mol}{L}V_{0}=0.18mol$

$V_{0}=\frac{0.18mol}{10.0\frac{mol}{L}}=0.018L$

V₀=18.0ml

The volume I have to take from 10.0M HCl solution is 18.0ml