A1.00 l buffer solution is 0.250 m in hf and 0.250 m in naf. calculate the ph of the solution after the addition of 100.0 ml of 1.00 m hcl. the ka for hf is 3.5 × 10-4.

pH = 3.09

Explanation:

Solve this problem using the Henderson-Hasselbach formula  after calculating the changes in the buffer following the addition of  HCl. The addition of the acid consumes the conjugate base in a one to one reaction. The moles present are calculated by multiplying the molarity by the volume in liters.

R:      HF                          ⇆   H⁺            +       F⁻

I:  0.250 mol                                             0.250  mol

C:+ 0.100 mol                                           -0.100   mol

E:  0.350   mol                                           0.150   mol

pH = pKa + log ( (F⁻) /(HF))

pH = -log ( 3.5 x 10 ⁻⁴)  + log ( 0.15/ 0.35)

pH =   3.46   +  ( -0.37 ) = 3.09

Note: Here we do not need to calculate the change in concentrations since the volume appears in the ratio and cancel themselves.

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.

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%