Determine the [oh-], ph, and poh of a solution with a [h+] of 0.090 m at 25 °c. [oh-] = ph = poh = 0 poh =

Answer : The concentration of ion, pH and pOH of solution is, , 1.05 and 12.95 respectively.

Explanation : Given,

Concentration of ion = 0.090 M

pH : It is defined as the negative logarithm of hydrogen ion or hydronium ion concentration.

The expression used for pH is:

First we have to calculate the pH.

The pH of the solution is, 1.05

Now we have to calculate the pOH.

The pOH of the solution is, 12.95

Now we have to calculate the concentration.

The concentration is,

pH = 1.046

[OH⁻] = 1.11 ×10⁻¹³ M

pOH = 12.954

Explanation:

Given: [H⁺] = 0.090 M = 9 ×10⁻² M;  T = 25°C  

As, pH = - log [H⁺]

⇒ pH = - log (9 ×10⁻²) = 1.046

The self-ionisation constant of water is given by ,

Kw = [H⁺] [OH⁻]

and, pKw = pH + pOH

Since at room temperature, 25°C: Kw = 1.0 × 10⁻¹⁴ , pKw = 14

∴ Kw = [H⁺] [OH⁻] = 1.0 × 10⁻¹⁴

⇒ [OH⁻] = (1.0 × 10⁻¹⁴) ÷ [H⁺] = (1.0 ×10⁻¹⁴)  ÷ [9 ×10⁻²] = 0.111 ×10⁻¹² = 1.11 ×10⁻¹³ M

And,

pH + pOH = pKw = 14

⇒ pOH = 14 - pH = 14 - 1.046 = 12.954

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error

it takes on an electronical energy