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Tris (hydroxymethyl)aminomethane, commonly called TRIS or Trizma, is often used as a buffer in biochemical studies. Its buffering range is from pH 7 to 9,9 , and KbK _ { b } is 1.19×1061.19 \times 10 ^ { - 6 } for the reaction

(HOCH2)3CNH2(aq)+H2O(l)(HOCH2)3CNH3+(aq)+OH(aq)TRISTRISH+\begin{array} {l}{{\left( \mathrm { HOCH } _ { 2 } \right) _ { 3 } \mathrm { CNH } _ { 2 } ( a q ) + \mathrm { H } _ { 2 } \mathrm { O } ( l ) \rightleftharpoons \left( \mathrm { HOCH } _ { 2 } \right) _ { 3 } \mathrm { CNH } _ { 3 } ^ { + } ( a q ) + \mathrm { OH } ^ { - } ( a q )} \\\quad {\text{TRIS}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \text{TRISH}^+}} \end{array}

A buffer is prepared by diluting 50.0 g of TRIS base and 65.0 g of TRIS hydrochloride (written as TRISHCl to a total volume of 2.0 L. What is the pH of this buffer? What is the pH after 0.50 mL of 12 M HCl is added to a 200.0-mL portion of the buffer?

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\bullet We have 50.0 g of TRIS.

The molar mass of TRIS is 121.14 g/mol, hence, the number of moles of TRIS is

nTRIS=50.0 g121.14 g/mol=0.41 mol\mathrm{ n_{TRIS} = \frac { 50.0\ g } { 121.14\ g/mol } = 0.41\ mol }

\bullet And we also have 65.0 g of TRISHCl.

The molar mass of TRISHCl is 157.6 g/mol, hence, the number of moles of TRISHCl is

nTRISHCl=65.0 g157.6 g/mol=0.41 mol\mathrm{ n_{TRISHCl} = \frac { 65.0\ g } { 157.6\ g/mol } = 0.41\ mol }

Since TRISHCl dissociates completely into TRISH+^+ and Cl^-, the number of moles of TRISH+^+ is 0.41 mol.

\bullet The volume of a solution is 2.0 L.

Note that we have equal numbers of moles of TRIS and TRISH+^+, hence, their concentrations will be the same

[TRIS]=[TRISH+]=0.41 mol2.0 L=0.205 M\mathrm{ [TRIS] = [TRISH^+] = \frac { 0.41\ mol } { 2.0\ L } = 0.205\ M }

Let us calculate the pH of a given buffer, using Henderson-Hasselbalch equation

pH=pKa+log([TRIS][TRISH+])=log(Ka)+log(0.2050.205)=log(8.40109)+0=8.088.1\begin{align*} \mathrm{ pH } &= \mathrm{ pK_a + log\left( \frac { [TRIS] } { [TRISH^+] } \right) }\\ &= \mathrm{ -log(K_a) + log \left( \frac { 0.205 } { 0.205 } \right) }\\ &= \mathrm{ -log(8.40 \cdot 10^{-9}) + 0 }\\ &= 8.08\\ &\approx {\color{#4257b2}8.1} \end{align*}

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