Question

Question #79774

Answer 1

`[HO^-]=3.98xx10^-11*mol*L^-1`

Explanation:

`[H_3O^+]=(3.00xx10^-3*mol)/(12*L)=2.5xx10^-4*mol*L^-1`

`pH=-log_10[H_3O^+]=-log_10(2.5xx10^-4)=-(-3.60)=3.60`

Since `pH+pOH=14`, `pOH=10.40`,

And now we take anti-logs.......

`[HO^-]=10^(-10.4)*mol*L^-1=3.98xx10^-11*mol*L^-1`

Hydroxide ion concentration is low, because this is an acidic solution.