Question

In a 1.0x10^-4 M solution of HClO(aq), identify the relative molar amounts of these species:HClO, OH-, H3O+, OCl-, H2O

Answer 1

The Formula is:

Molar Mass of an Element = x Relative mass of atoms Molar mass constant (1g / mol)

Further Explanation

Understand molar mass

Molar mass = mass (in grams) of 1-mol substances

Molar mass is the mass (in grams) of one mole of a substance. By using the atomic mass of an element and multiplying it by the conversion factor of grams per mole (g / mol), you can calculate the molar mass of the element.

Look for the relative atomic mass of elements

Relative atomic mass:

Hydrogen = 1,007

Carbon = 12.0107

Oxygen = 15,9994

Chlorine = 35,453

The relative atomic mass of an element is the average mass of a sample of all its isotopes in atomic units. This information can be found in the periodic table of elements.

Multiply the atomic mass relative to the molar mass constant

Molar Mass of an Element = x Relative mass of atoms Molar mass constant (1g / mol)

Hydrogen = 1,007 x 1g / mol = 1,007 g / mol

Carbon = 12.0107 g / mol

Oxygen = 15,999 g / mol

Chlorine = 35,453 g / mol

H2: 1,007 x 2 = 2,014 g / mol

O2: 15,999 x 2 = 31,9988 g / mol

C2: 35,453 x 2 = 70,096 g / mol

example:

Hydrochloric acid

HCl -> 1 hydrogen atom, 1 chlorine atom

Glucose

C6H12O6 -> 6 carbon atoms, 12 hydrogen atoms, 6 oxygen atoms

Learn More

Molar mass brainly.com/question/2194946

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Details

Class: Middle/High School

Subject: Chemistry

Keywords: Mass, Molar, atoms

Answer 2

HClO is a weak acid, which means the ions do not fully dissociate. The hydrolysis reaction for the hypochlorous acid is:

HClO + H2O ⇄ H3O+ +OCl-

Then the equilibrium constant, Ka, of dilute HClO would be:

$K_{a} = \frac{[ H_{3} O^{+} ][O Cl^{-} ]}{HClO}$

Then we do the ICE table. I is for the initial concentration, C for the change and E for the excess.
      
          HClO       + H2O   ⇄   H3O+ +  OCl-
I     1.0x10^-4                          0             0
C        -x                                 +x           +x 
E  (1.0x10^-4 - x)                     x             x

Substituting the excess (E) concentration to the Ka equation:

$K_{a} = \frac{[x ][x]}{1.0 \ x \ 10^{-4} - x }$

Simplifying the equation would yield a quadratic equation:

$x^{2} + K_{a}x-(1.0 \ x \ 10^{-4}) K_{a}=0$

The Ka for HClO is an experimental data which was determined to be 2.9 x 10^-8. Substitute this to the equation, determine the roots, then you get the value for x, which is the concentration of H3O+ and ClO-. Just use your calculator feature Shift-Solve.

x = 1.688 x 10^-6 M = [H3O+] = [ClO-]

Then, you can determine the conc of [OH-] through pH.

pH = -log {H3O+] = -log [1.688 x 10^-6] = 5.77
pOH = 14 - pH = 14 - 5.77 = 8.23
pOH = 8.23 = -log [OH-]
[OH-] = 5.89 x 10^-9 M

Also, since HClO is (1.0x10^-4 - x), then it's concentration would be:
[HClO] = 1.0x10^-4 - 1.688 x 10^-6 = 9.83 x10^-5 M

Let's summarize all concentrations:
[HClO] = 9.83 x10^-5 M
[OH-] = 5.89 x 10^-9 M
[H3O+] = [ClO-] = 1.688 x 10^-6 M
Since the solution is dilute, H2O is relatively higher in concentration.

Thus in relative amounts, the order would be

H2O >>> HClO > H3O+ = ClO- > OH-