<u>Answer:</u> The initial molarity of cation is 2.38 M
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

We are given:
Given mass of
= 46.3 g
Molar mass of
= 129.6 g/mol
Volume of solution = 150 mL
Putting values in above equation, we get:

1 mole of
produces 1 mole of
cation and 1 mole of
anion
So, molarity of
cation = (1 × 2.38) = 2.38 M
Hence, the initial molarity of cation is 2.38 M
Remember pH is defined as: pH = -log[H+]
so, substituting
4 = - log[H+]
-4 = log [H+] Take antilog of both sides
10^(-4) = [H+]
[H+] = 1 x 10^-4 M
what is that????????????????