You are making some MES buffer. The pKa of MES is 6.2. You want to make 1 L of a 1 M solution of MES (pH 6.2). You weigh out 1 mole of the basic (unprotonated) form of MES and dissolve it in 750 mL water. What would you add to this before you brought the volume to 1 L with water?

To make the buffer you must add 1 mole of MES in the liter of the buffer.

Explanation:

To find pH of the buffer of MES you must use Henderson-Hasselbalch equation:

pH = pKa + log [MES⁻] / [MES]

Where [MES⁻] is molar concentration of unprotoned MES and [MES] molar concentration of the acid

Replacing:

6.2 = 6.2 + log [1M] / [MES]

Molar concentration of [MES⁻] is 1mol / L = 1M

0 = log [1M] / [MES]

1 = [1M] / [MES]

[MES] = 1M

That means to make the buffer you must add 1 mole of MES in the liter of the buffer

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%