5. what concentration of acid must be added to change the ph of 1 mm phosphate buffer from 7.4 to 7.3 (pkas of the phosphate buffers are 2.12, 7.21 and 12.67)?

Explanation:

According to the Henderson-Hasselbalch equation, the relation between pH and is as follows.

               pH =

where,     pH = 7.4 and = 7.21

As here, we can use the nearest to the desired pH.

So,      7.4 = 7.21 +

             0.19 =

            = 1.55

1 mM phosphate buffer means + = 1 mM

Therefore, the two equations will be as follows.

            = 1.55 ............. (1)

  + = 1 mM ........... (2)        

Now, putting the value of from equation (1) into equation (2) as follows.

             1.55 = 1 mM

                        2.55 = 1 mM

                              = 0.392 mM

Putting the value of in equation (1) we get the following.

                     0.392 mM + = 1 mM

                          = (1 - 0.392) mM

                              = 0.608 mM

Thus, we can conclude that concentration of the acid must be 0.608 mM.

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%