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

`pH=7.72`

Step-by-step 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(([TRIS])/([TRIS-HCl]) )`

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

`[TRIS]=(7.0g/(121.14g/mol))/(0.250L)=0.13M`

`[TRIS-HCl]=(12.0g/(157.60g/mol))/(0.250L)=0.305M`

Then, the pH:

`pH=8.08+log((0.13M)/(0.305M) )\\\\pH=7.72`

Best regards!