A0.001 molar solution of weak acid has a ph = 4.0. what is the pka of the acid?
a. 4
b. 5
c. 6
d. 7
e. 8

A 0.001 molar solution of weak acid has a pH = 4.0,The pKa is :

B. 5

Explanation:

pH = It is negative logarithm of hydrogen ion concentration.

pH = -log([H+])

Ka = It is the dissociation constant.

pKa = -log(Ka)

The acid dissociates according to the equation given below:

Here , The concentration of A- and H+ will be same

So,

[H+] = [A-].....(1)

Now ,

pH = 4.0

pH = -log([H+])

4.0 = -log([H+])

- 4.0 = log([H+])        take antilog both side::"use scientific calculator"

From equation (1), [A-] =

[HA] = 0.001 M (given)

Insert the value of HA , H+ and A- in Ka

pKa = -log(Ka)

pKa = 5

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%