Question

What is the ph of a solution of 1.0 × 10-9 m naoh (include 1 digit after the decimal in your answer)?

Answer 1

Answer is: pH of NaOH solution is 5.
Reaction: NaOH → Na⁺ + OH⁻.
c(NaOH) = 1,0·10⁻⁹ mol/dm³.
pH = ?
from reaction: c(OH⁻) = c(NaOH).
c(OH⁻) = 1,0·10⁻²³ mol/dm³.
pOH = -log(c(OH⁻)) = 9.
pH + pOH = 14.
pH = 14 - pOH.
pH = 14 - 9 = 5.
pH is potential of hydrogen.

Answer 2

Answer:

$pH=5$

Explanation:

Hello,

In this case, for the dissociation of the sodium hydroxide into sodium and hydroxyl ions we have:

$NaOH\rightarrow Na^++OH^-$

Which is a 100% dissociation as sodium hydroxide is a strong base, therefore, the concentration of the hydroxyl ions are 1.0x10⁻⁹. For that reason, the pOH could be computed as:

$pOH=-log([OH^-})=-log(1.0x10^{-9})=9$

Finally, from the definition of pH, we have:

$pH+pOH=14\\pH=14-pOH=14-9\\pH=5$

Best regards.