What is the concentration of hydroxide ions in pure water at 30.0°C, if #K_w# at this temperature is #1.47*10^-14#?

1 Answer
Feb 12, 2017

Answer:

We interrogate the equilibrium: #"2H_2O(l) rightleftharpoons H_3O^(+) + HO^-#.

#[HO^-]=1.21xx10^-7*mol*L^-1#

Explanation:

We interrogate the equilibrium:

#"2H_2O(l) rightleftharpoons H_3O^(+) + HO^-#,

where the ion product, #K_w=1.47xx10^-14#, at #303*K#. Note that this is slightly higher than #K_w=10^-14# at #298*K#, and given that this is a bond breaking reaction this is perhaps reasonable.

From the equation, #[H_3O^+][HO^-]=1.47xx10^-14#.

Given that this is a neutral solution, #[H_3O^+]=[HO^-]#.

Thus #[HO^-]^2=1.47xx10^-14#

And #[HO^-]=sqrt(1.47xx10^-14)=1.21xx10^-7*mol*L^-1#