a pH meter gives a readout of 9.35 when the probe is dipped into an aqueous solution containing the strong base NaOH. What is the molarity of this solution with the respect to sodium hydroxide?
pH = -log(H^+)
9.35 = -log(H^+).
solve for )H^+). Do you know how to do that on the calculator. Punch in 9.35, change the sign to negative, hit the 10x key. The (H^+) pops up
Then (H^+)(OH^-) = Kw.
You know H and Kw. Solve for (OH^-)
There is an easier way, I think, to do it.
9.35 is pH. Then pOH = 14-9.35.
Then pOH = -log(OH^-).
Substitute for pOH, and solve for OH^-. Much simpler, I think.
Check my work.
ok since 14-9.35=4.65
what do i do after this...
pH = -log(OH^-)
4.65 = -log(OH^-)
Just follow the above instructions to convert this number to OH^-