Asked by Cliff
Hi,
I'm trying to solve some chemistry problems, like the ones my teacher says will appear on our exams.
The problem reads like this:
What is the hydrogen conentration in a bleach sample that register 9.55 using a pH meter?
My Attempt:
Since this sample register a 9.55 on a pH meter,it occurs to me that this sample is basic, and I can use the formula:
pOH = -log (H^+), to solve for (OH^-)
=-log(9.55)
=-0.980
From here I thought,maybe if I substitute that result into the equation pH + pOH = 14, I might arrive at the answer I'm looking for; which is 0 000 000 000 35 M.
pH + 0.980 = 14
pH = 13.02
Evidently I took the wrong approach.
Can you help me with this problem?
Have a Great Night
I'm trying to solve some chemistry problems, like the ones my teacher says will appear on our exams.
The problem reads like this:
What is the hydrogen conentration in a bleach sample that register 9.55 using a pH meter?
My Attempt:
Since this sample register a 9.55 on a pH meter,it occurs to me that this sample is basic, and I can use the formula:
pOH = -log (H^+), to solve for (OH^-)
=-log(9.55)
=-0.980
From here I thought,maybe if I substitute that result into the equation pH + pOH = 14, I might arrive at the answer I'm looking for; which is 0 000 000 000 35 M.
pH + 0.980 = 14
pH = 13.02
Evidently I took the wrong approach.
Can you help me with this problem?
Have a Great Night
Answers
Answered by
pranav
i did not exactly get that answer... but i still think this will helps. i did 10^pH formula. i got 2.818 x 10^-10 as my naswer. i toke that and divided it by the molar mass of blaech. i got the answer 3.78 x 10^12... which is simalr to your answer
Answered by
DrBob222
I think you are trying to make this far more difficult that it is. Also, I think your prof gave you the wrong answer (or you've state something wrong or copied something wrong).
if pH = 9.55, then
pH = -log(H^+)
9.55 = -log)H^+)
-9.55 = log(H^+)
(H^+) = 2.82E-9M
(the OH^- is 3.5E-5M
(You should know your answer of pH = 13.02 isn't right. the PROBLEM states pH = 9.55 and not 13.02.
By the way, pOH = -log(OH^-), not what you wrote above.
if pH = 9.55, then
pH = -log(H^+)
9.55 = -log)H^+)
-9.55 = log(H^+)
(H^+) = 2.82E-9M
(the OH^- is 3.5E-5M
(You should know your answer of pH = 13.02 isn't right. the PROBLEM states pH = 9.55 and not 13.02.
By the way, pOH = -log(OH^-), not what you wrote above.
Answered by
DrBob222
And my problem is I can't read my calculator. For pH = 9.55, (H^+) = 2.82E-10.
Answered by
Cliff
I submitted an incorrect answer to that problem. The correct answer is 0.000 000 000 28 M. I apolgise.
I've been sleep deprived.
I've been sleep deprived.
Answered by
DrBob222
So the answer 2.8E-10 is correct. I should have rounded that last two out of the number and didn't.
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