Question

What mass of HI should be present in 0.300L of solution to obtain a solution the following pH?

pH=1.75

Chemistry

Answer 1

HI + H₂O -------> H₃O⁺ + I^-

HI is a strong acid which completely dissociates in water

pH = -log[H⁺]

[H⁺] = [H₃O⁺]

[H₃O⁺] = 10^[-pH]   

= 10^[-1.75]

[H₃O⁺] = 0.017783 M

Molarity = moles of solute/litre of solution

0.017783 M = moles of [H₃O⁺] / 0.300 L

moles of [H₃O⁺] = 0.017783*0.300

              = 0.005335 mol

Molecular weight of HI = 127.911 g/mol

Mass of HI = (0.005335 mol)* (127.911 g/mol) = 0.6824 g

Mass of HI present in the solution = 0.6824 g