The pH of a solution changes fro 300 to 6.00. By what factor does the $[H_3O^+]$ change?

Question

5465 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-26T16:40:08

By a factor of $1000$ .

$pH=-log_10[H_3O^+]$

If $pH_1=3.00$ , then $[H_3O^+]=10^-3*mol*L^-1.$

If $pH_2=6.00$ , then $[H_3O^+]=10^-6*mol*L^-1.$

$(pH_1)/(pH_2)=(10^-3*mol*L^-1)/(10^-6*mol*L^-1)=10^3$

The latter concentration is less concentrated with respect to $H_3O^+$ than the former by a factor of $1000$ .

The pH of a solution changes fro 300 to 6.00. By what factor does the [H_3O^+][H_3O^+] change?