Question

Calculate the concentrations of hydronium and hydroxide ions in a solution prepared by dissolving 0.798 g of HCl gas in enough water to make 443 mL of solution.

[OH−] =

[H3O+] =

Answer 1

Molar mass of HCl = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass of HCl = 0.798 g

we have below equation to be used:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(0.798 g)/(36.458 g/mol)

= 2.189*10^-2 mol

volume , V = 443 mL

= 0.443 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 2.189*10^-2/0.443

= 4.94*10^-2 M

[H3O+] = [HCl] = 4.94*10^-2 M

we have below equation to be used:

[OH-] = Kw/[H3O+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(4.94*10^-2)

[OH-] = 2.02*10^-13 M

[OH-] = 2.02*10^-13 M

[H3O+] = 4.94*10^-2 M