Q163CP - Chemistry: Molecular Nature Of Matter And Change

Question

Tris(hydroxymethyl)aminomethane, known as TRIS or THAM, is a water-soluble base used to synthesize surfactants and pharmaceuticals, as an emulsifying agent in cosmetics, and in cleaning mixtures for textiles and leather. In biomedical research, solutions of TRIS are used to maintain nearly constant pH for the study of enzymes and other cellular components. Given that the pKb is 5.91, calculate the pH of 0.075 M TRIS.

Step-by-Step Solution

Answer

The pH of 0.075 M TRIS is 10.48.

Step 1: Calculate the pH of 0.075 M TRIS

Note that TRIS is a water-soluble base. Therefore, it produces ${\text{OH}}^{-}$. Write the simplified reaction equation.

$\text{TRIS}+{\text{H}}_2{\text{O}}^{+}\to {\text{TRISH}}^{\text{ + }}+{\text{OH}}^{-}$

Next, construct the ICE table to obtain the equation for ${\text{K}}_{\text{b}}$.

$$\begin{gathered} {\text{K}}_{\text{b}}\text{ = }\frac{\left[{\text{OH}}^{\text{ - }}\right]\left[{\text{TRISH}}^{\text{ + }}\right]}{\text{[TRIS]}} \\ \text{ = }\frac{{\text{x}}^{\text{2}}}{\text{0.075 - x}}. \end{gathered}$$

First, solve for the ${\text{K}}_{\text{b}}$  from the given ${\text{pK}}_{\text{b}}.$

$\begin{array}{rcl}& & {\text{pK}}_{\text{b}}\text{ = } \text{ - log}\left({\text{K}}_{\text{b}}\right)\\ & & {\text{ K}}_{\text{b}}\text{ = } {\text{10}}^{{\text{ - pK}}_{\text{b}}}\\ & & \text{ = } {\text{10}}^{\text{ - 5.91}}\\ & & \text{ = } \text{1.23}\times {\text{10}}^{\text{ - 6}}.\end{array}$

Step 2: Replace the values

We know that  $\text{x} \text{ = } \left[{\text{TRISH}}^{\text{ + }}\right] \text{ = } \left[{\text{OH}}^{\text{ - }}\right]$. Since TRIS is a weak base, its ${\text{K}}_{\text{b}}$ must be very small. So, assume that the x has no effect on  in the denominator. Replace the ${\text{K}}_{\text{b}}$  to solve for X.

$\begin{array}{rcl}& & {\text{K}}_{\text{b}}\text{ = }\frac{{\text{x}}^{\text{2}}}{\text{0.075}}\\ & & {\text{x}}^{\text{2}}\text{ = }\left({\text{K}}_{\text{b}}\right)\text{(0.075)}\\ & & \text{ x = }\sqrt{\left({\text{K}}_{\text{b}}\right)\text{(0.075)}}\\ & & \text{ = }\sqrt{\left(\text{1.23}\times {\text{10}}^{\text{ - 6}}\right)\text{(0.075)}}\\ & & \text{ = } \text{3.04}\times {\text{10}}^{\text{ - 4}}.\end{array}$

Since $\text{x} \text{ = } \left[{\text{TRISH}}^{\text{ + }}\right] \text{ = } \left[{\text{OH}}^{\text{ - }}\right]\text{, then }\left[{\text{OH}}^{\text{ - }}\right] \text{ = } \text{3.04}\times {\text{10}}^{\text{ - 4}}\text{M.}$

Next, calculate the pOH of the solution.

$\begin{array}{rcl}& & \text{pOH = } \text{ - log}\left[{\text{OH}}^{\text{ - }}\right]\\ & & \text{ = } \text{ - log}\left(\text{3.04}\times {\text{10}}^{\text{ - 4}}\right)\\ & & \text{ = } \text{3.52.}\end{array}$

Lastly, convert the pOH to pH.

$\begin{array}{rcl}& & \text{pH + pOH = } \text{14}\\ & & \text{pH} \text{ = } \text{14 - pOH}\\ & & \text{ = } \text{14 - 3.52}\\ & & \text{ = } \text{10.48.}\end{array}$

Hence, the pH of 0.075 M TRIS is 10.48.