## Related questions with answers

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 $9 ,$ and $K _ { b }$ is $1.19 \times 10 ^ { - 6 }$ for the reaction

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

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?

Solution

Verified#### (c)

$\bullet$ 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

$\mathrm{ n_{TRIS} = \frac { 50.0\ g } { 121.14\ g/mol } = 0.41\ mol }$

$\bullet$ 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

$\mathrm{ n_{TRISHCl} = \frac { 65.0\ g } { 157.6\ g/mol } = 0.41\ mol }$

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

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

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

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

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