Calculate the pH of a 0.078 M solution of NaOCl (Ka of HClO = 3.5 x 10-8). Enter pH to 3 decimal places.

The pH of the solution is 4.282

Explanation:

First we need to write the dissociation reaction

 ⇄

Also, we need to prepare I C E table

             ⇄

I          0.078                               0             0

C         - x                                    +x            +x

E        0.078 - x                            x               x

Thus, pH= - Log[H₃O⁺]

The equilibrium concentration of  H₃O⁺ = x

So, we solve for x

Solving this quadratic equation; x = 5.2232 x 10⁻⁵ M

pH = - Log[H₃O⁺]

pH = - Log[5.2232 x 10⁻⁵]

pH = 4.282 (in 3 decimal places)

Therefore, the pH of the solution is 4.282

first, consult an activity series chart to determine if al is more reactive than fe. it is, so it will replace the fe. now, one can continue with the problem.

next, make sure you have a correctly balanced equation.

3feo + 2al ==> 3fe + al2o3

now, determine which reactant (feo or al) is in limiting supply.

moles feo present = 125 g x 1 mole/71.8 g = 1.74 moles

moles al present = 25.0 g x 1 mole/26.98 g = 0.927 moles

since from the balanced equation, it take 3 mol feo for every 2 mol al, al will be limiting.

you can determine this simply by dividing moles of each by their coefficient in the balanced equation. for example: 1.74/3 = 0.58 and 0.927/2 = 0.464. al is less, so it is limiting.

finally, using the stoichiometry in the balanced eq. determine mass of fe that can be produced from 0.927 moles of al.

0.927 moles al x 3 moles fe/2 moles al x 55.8 g/mole = 77.6 g

it is to be noted that an alpha particle is an helium nucleus. so it contains 2 protons 2 neutrons and has a mass number of 4. when alpha decay particles are ejected from the nucleus it gets transformed into an atom which has atomic number less than 2 and a mass number less than 4.

an equation that could represent alpha decay is radium 226 decay of alpha particle forms radon 222.

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