Question

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

localid="1664876042279" $\begin{array}{l}CN{H}_2\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons CN{H_3}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)\\ TRISTRIS{H}^{+}\end{array}$

a.What is the optimum pH for TRIS buffers?

b.Calculate the ratio $\left[TRIS\right]/\left[TRIS{H}^{+}\right]$at$pH=7.00$and at$pH=9.00$.

c.A buffer is prepared by diluting 50.0 g of TRIS base and 65.0 g of TRIS hydrochloride 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?

Short answer

a. The optimum pH is 8.08.

b. The ratio of $\left[TRIS\right]/\left[TRIS{H}^{+}\right]$at pH 7 is 0.083.

The ratio of $\left[TRIS\right]/\left[TRIS{H}^{+}\right]$at pH 9 is 8.40.

c. The pH of this buffer solution is 8.08.

The pH after 0.50 mL of 12 M HCl is added to a 200.0 mL portion of the buffer is 7.95.

Step-by-step solution

Subpart (a) The optimum pH of a buffer occurs when.pH=pKa

$\begin{array}{l}{K}_a=8.40\times {10}^{-9}\\ p{K}_a=-log{K}_a\\ p{K}_a=8.08\\ pH=8.08\end{array}$

Subpart (a) The ratio  of [TRIS]/[TRISH+] at different pH.

The Henderson Hesselbalch equation

$\begin{array}{l}pH=p{K}_a+log\frac{\left[conjugatebase\right]}{\left[acid\right]}\\ pH=p{K}_a+log\frac{\left[TRIS\right]}{\left[TRIS{H}^{+}\right]}\end{array}$

At pH 7,

$\begin{array}{l}7.00=8.08+log\frac{\left[TRIS\right]}{\left[TRIS{H}^{+}\right]}\\ \frac{\left[TRIS\right]}{\left[TRIS{H}^{+}\right]}={10}^{-1.08}\\ \frac{\left[TRIS\right]}{\left[TRIS{H}^{+}\right]}=0.083\end{array}$

At pH 9,

$\begin{array}{l}log\frac{\left[TRIS\right]}{\left[TRIS{H}^{+}\right]}=9.00-8.08\\ \frac{\left[TRIS\right]}{\left[TRIS{H}^{+}\right]}={10}^{0.924}\\ \frac{\left[TRIS\right]}{\left[TRIS{H}^{+}\right]}=8.40\end{array}$

 Step 3: Subpart (c) The pH.

Using Henderson Hesselbalch equation,

$\begin{array}{l}m\left(TRIS\right)=50.0g\\ n\left(TRIS\right)=0.413mol\\ m\left(TRISHCl\right)=65.0g\\ Mw\left(TRISHCl\right)=157.60gmo{l}^{-}\\ n\left(TRISHCl\right)=0.412mol\end{array}$

Volume of solution V = 2.0L

And,

The concentration,

$\begin{array}{l}\left[TRIS\right]=\frac{numberofmoles}{volumeofsolutioninlitres}\\ =0.413mol/2.0L\\ =0.206mol/L\end{array}$

By Henderson Hesselbalch equation to find pH,

$\begin{array}{l}pH=p{K}_a+log\frac{\left[TRIS\right]}{\left[TRIS{H}^{+}\right]}\\ pH=8.08+log\frac{0.206}{0.206}\\ pH=8.08\end{array}$

The pH of this buffer solution is 8.08.

In 0.50 mL of 12MHCl,

$\begin{array}{l}n\left(HCl\right)=0.50mL\times {10}^{-3}L/mL\times 12M\\ n\left(HCl\right)=0.006mol\end{array}$

New concentrations of TRIS and TRISHCl is

$\begin{array}{l}\left[TRIS\right]\text{'}=0.176M\\ \left[TRISHCl\right]=0.235M\\ pH=8.08+log(0.176/0.235)\\ pH=7.95\end{array}$

The pH after 0.50 mL of 12 M HCl is added to a 200.0 mL portion of the buffer is $7.95$.