What is (oh-) in a solution of 0.125m ch3nh2 and 0.130m ch3nh3cl? Ka = 4.4×10^-4

Explanation:

The solution which contains 0.125 M CH3NH2 and 0.130 M CH3NH3Cl acts as buffer solution and the pOH of the buffer solution use the following Henderson-

Hasselbalch equation:

pOH = pKo + log(BH+)/[B]

Given that

CH3NH2 (aq) + H2O(l) + CH3NH, (aq) + OH- (aq)

Ky = 4.4 x 10-4

Here, the value of Ky

=- 4.4 x 10-4 ,Therefore pK is 3

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%

a is the correct answer