Question

What is the pH of a solution prepared by dissolving 7.0 g of tris (FM 121.14) plus 12.0 g of tris hydrochloride (FM 157.60) in 250 mL water

Answer 1

Answer:

$pH=7.72$

Explanation:

Hello,

In this case, considering that TRIS is a weak base and the TRIS hydrochloride its conjugate acid with a pKa of about 8.08, we can use the Henderson-Hasselbach equation in order to compute the pH of the described solution:

$pH=pKa+log(\frac{[TRIS]}{[TRIS-HCl]} )$

Next, we compute the concentrations of TRIS and TRIS hydrochloride as shown below (mol/L):

$[TRIS]=\frac{7.0g/(121.14g/mol)}{0.250L}=0.13M$

$[TRIS-HCl]=\frac{12.0g/(157.60g/mol)}{0.250L}=0.305M$

Then, the pH:

$pH=8.08+log(\frac{0.13M}{0.305M} )\\\\pH=7.72$

Best regards!