What weight of HCl is required in a 0.42 solution, calculate the Ka and the pH of the solution.?

Question

ka= multiplication of concentration of products / multiplication of concentration of reactant.
For example A+B give us C+D
Ka= [A] * [B] /[C]*[D]

Answer 1

Assuming you mean 0.42 molar solution.
HCl has a relative molecule mass of 36.5, so by
Moles = mass / molar mass
0.42 = mass / 36.5
mass = 36.5 * 0.42 = 15.33g
So you'll need 15.33 grams of HCl per litre of water.
Ka is the acid dissociation constant, which is always the same for HCl, at 10^8. This is not given by the equation you use - firstly you've got it upside down (your word description is correct however), and secondly when it comes to acids, there is no B.
pH = -log[H+]
[H+] = 0.42 as there is one H+ per HCl (if HCl were a weak acid we'd use the Ka equation Ka=([H+]^2)/[HA], as it would not fully ionise)
so pH ~= 0.377

Answer 2

I think you are missing some words...can you please re-read and write again?