Question

Determine the (H+), pH, and pOH of a solution with an [OH-] of 9.5 x 10-10 M at 25 °C. M pH =

Answer 1

pOH = 9.022, [H⁺] = 1.5×10⁻⁵ M, pH = 4.978

Step-by-step explanation:

Given: [OH⁻] = 9.5 × 10⁻¹⁰ M, T= 25°C

As, pOH = - log [OH⁻]

⇒ pOH = - log (9.5 x 10⁻¹⁰) = 9.022

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 and pKw = 14.

Therefore, Kw = [H⁺] [OH⁻] = 1.0 × 10⁻¹⁴

⇒ [H⁺] = (1.0 × 10⁻¹⁴) ÷ [OH⁻] = (1.0 ×10⁻¹⁴) ÷ [9.5 × 10⁻¹⁰] = 0.105 ×10⁻⁴ = 1.5×10⁻⁵ M

also,

pH + pOH = pKw = 14

⇒ pH = 14 - pOH = 14 - 9.022 = 4.978

Answer 2

Answer : The concentration of
`H^+` ion, pH and pOH of solution is,
`1.05* 10^(-5)M`, 4.98 and 9.02 respectively.

Explanation : Given,

Concentration of
`OH^-` ion =
`9.5* 10^(-10)M`

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

The expression used for pH is:

`pH=-\log [H^+]`

First we have to calculate the pH.

`pOH=-\log [OH^-]`

`pOH=-\log (9.5* 10^(-10))`

`pOH=9.02`

The pOH of the solution is, 9.02

Now we have to calculate the pH.

`pH+pOH=14\\\\pH=14-pOH\\\\pH=14-9.02=4.98`

The pH of the solution is, 4.98

Now we have to calculate the
`H^+` concentration.

`pH=-\log [H^+]`

`4.98=-\log [H^+]`

`[H^+]=1.05* 10^(-5)M`

The
`H^+` concentration is,
`1.05* 10^(-5)M`