Determine the ph of a solution that is 1.35% naoh by mass. assume that the solution has density of 1.01 g/ml.

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13.545

Explanation: -

Let the volume of the solution be 1000 mL

Density = 1.04 g/ mL

Mass = density x vol

= 1.04 g/ mL x 1000 mL

= 1040 g

Amount of NaOH = x 1040 g

= 14.04 g

Molar mass of NaOH = 23 x 1 + 16 x 1 + 1 x 1

= 40 g / mol

Number of moles = Amount of NaOH / Molar mass

= 14.04 /40

= 0.351 mol

So 1000 ml of the solution has 0.351 mol

Molarity of NaOH = 0.351 M

Using the relation

pOH = - log [OH-]

= - log 0.351

= 0.455

Using the relation pH + pOH = 14

pH = 14 - pOH

= 14 - 0.455

= 13.545

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.