Question

How many grams of Tris base (Trizma base) and sodium chloride are weighed out to make one liter of 10x Tris-buffered saline, pH 7.4 (TBS) which contains 100mM Tris base and 1.5 M NaCl?

Answer 1

Ans. #1. Amount of Tris base required = 100 mM

                                                            = 0.1 M                                               ; [1 M = 10³ mM]

                                                            = 0.1 mol/ L                                       ; [1 M = 1 mol/ L]

                                                            = (0.1 mol x 121.14 g mol^-1) / L

                                                            = 12.114 g/ L

The above quantity required is for 1X Tris base.

Note that 10X means that the solution is 10 times concentrated than the 1X (= 100 mM) solution.

So,

            Amount required = 10 times of 1X

                                                = 10 x (12.114 g/L)

                                                = 121.14 g/ L

Thus, the amount of Tris base required for 1 L solution = 121.14 g

#2. Amount of NaCl required = 1.5 M

                                                            = 1.5 mol/ L                                      

                                                            = (1.5 mol x 58.44 g mol^-1) / L

                                                            = 90.582 g/ L

The above quantity required is for 1X NaCl solution.

Note that 10X means that the solution is 10 times concentrated than the 1X (= 1.5 M) solution.

So,

            Amount required = 10 times of 1X

                                                = 10 x (90.582 g/L)

                                                = 905.82 g/ L

Thus, the amount of NaCl required for 1 L solution = 905.82 g

Note: The concentration of NaCl appear to be very high. Please check if its concentration is 1.5 M.