Calculate the [oh-] and the ph of a solution with an [h+= .00083 m at 25 degrees

1.2 × 10^-11 M

We are given;

Concentration of Hydrogen ions [H⁺] as 0.00083 M

We are required to calculate the concentration of [OH⁻]

We know that;

pH = - log [H⁺]

POH = -log[OH⁻]

Also, pH + pOH = 14

With the concentration of H⁺ we can calculate the pH

pH = -log 0.00083 M

    = 3.08

But; pH + pOH = 14

Therefore; pOH = 14 - pH

                           = 14 - 3.08

                          = 10.92

But, pOH = -log[OH⁻]

Therefore; [OH] = -Antilog(pOH)

Hence; [OH⁻] = Antilog -10.92

                     = 1.2 × 10^-11 M

Therefore, [OH⁻] is 1.2 × 10^-11 M

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o