Question

Part C

a solution that is 4.0×10⁻² M in HClO4 and 5.3×10⁻² M in HCl

Express your answer to two decimal places.

pH =

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Part D

a solution that is 1.11% HCl by mass (Assume a density of 1.01 g/mL for the solution.)

Express your answer to three decimal places.

pH =

Answer 1

Solution :-

Part c

4.0*10^-2 M HClO4 and 5.3*10^-2 M HCl

Both are strong acids therefore total H+ concentration is the sum of the concentrations of the both acids

Total [H+] = 4.0*10^-2 M + 5.3*10^-2 M = 9.3*10^-2 M

pH= -log [H^+]

    = - log [ 9.3*10^-2]

= 1.03

Part d

A solution with 1.11% HCl by mass and density is 1.01 g/ml

Lets assume we have 1 L solution

Then mass of solution = (1 L *1000 ml / 1 L)*1.01 g per ml = 1010 g

Mass of HCl = (1010 g solution * 1.11 % HCl) / 100 %

                      = 10.201 g

Moles of HCl = mass/ molar mass

                         = 10.201 g / 36.5 g per mol

                         = 0.279 mol HCl

Molarity of HCl = mole / volume

                             = 0.279 mol / 1 L

                          = 0.279 M

pH= -log [H^+]

pH= -log [0.279]

pH= 0.554