Question

Determine the pH of each of the following solutions., 3.6×10−2 M HI,9.23×10−2 M HClO4, a solution that is 4.0×10−2 M in HClO4 and 4.8×10−2 M in HCl, a solution that is 1.01% HCl by mass (Assume a density of 1.01 g/mL for the solution.)

please show work

Answer 1

Answer – We are given the acids and need to calculate the pH of each

1)[HI] = 3.6*10^-2 M

We know the HI is the strong acid, so

[HI] = [H₃O⁺] = 3.6*10^-2 M

We know,

pH = -log [H₃O⁺]

      = -log 3.6*10^-2 M

     = 1.44

2)[HClO₄] = 9.23*10^-2 M

We know the HClO₄ is the strong acid, so

[HClO₄] = [H₃O⁺] = 9.23*10^-2 M

We know,

pH = -log [H₃O⁺]

      = -log 9.23*10^-2 M

     = 1.03

3) Solution [HClO₄] = 4.0*10^-2 M and [HCl] = 9.23*10^-2 M

We know the HClO₄ and HCl both are the strong acid, so

[HClO₄] = [H₃O⁺] = 4.0*10^-2 M

[HClO₄] = [H₃O⁺] = 4.8*10^-2 M

Total, [H₃O⁺] = 4.0*10^-2 M + 4.8*10^-2 M

                       = 8.8*10^-2 M

We know,

pH = -log [H₃O⁺]

      = -log 8.8*10^-2 M

     = 1.05

4) solution that is 1.01% HCl by mass, density of solution = 1.01 g/mL

We assume 100 g of solution

So, mass of HCl = 1.01 g

volume of solution = 100 g / 1.01g.mL^-1

                                 = 99.0 mL

                                = 0.099 L

Moles of HCl = 1.01 g / 36.456 g.mol^-1

                        = 0.0277 moles

[HCl] = 0.0277 moles / 0.099 L

          = 0.280 M

We know,

HCl is the strong acid, so

[HClO₄] = [H₃O⁺] = 0.280M

We know,

pH = -log [H₃O⁺]

      = -log 0.280 M

     = 0.553