pH meter/ calculation?

[evusq]

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Please help! I've been working on this problem for so long with the solution right in front of me and I still don't understand- I think it's more of a conceptual/problem solving thing than chem... but any help would be appreciated!

So I'm given the following equation: Eobs= E'cell- .0592*log(H+)
(E'= standard) The pH meter measures Eobs to calculate the pH.
The value for E' is given as 0.285V.

Q: If the E' were actually .3V, instaed of .285V, what would be true of the pH measured by the pH meter, assuming no correction is made?

The answer says that the pH determined will be too low, because E' in calculation is too high.

First of all, isn't the E' used in calculation too low, because the wrong value is used? And also, if E'used is smaller than E'actual, wouldn't the second log term also be smaller, so H+ is lower, and pH is higher? I rewrote the equation above as Eobs= E' + .592pH, in which case the pH term should be bigger for smaller E'.

I'm not sure where I'm stuck.. please feel free to correct me where I'm wrong and any extra explanation would be very helpful. Thanks so much~

 

[ezsanche]

Please help! I've been working on this problem for so long with the solution right in front of me and I still don't understand- I think it's more of a conceptual/problem solving thing than chem... but any help would be appreciated!

So I'm given the following equation: Eobs= E'cell- .0592*log(H+)
(E'= standard) The pH meter measures Eobs to calculate the pH.
The value for E' is given as 0.285V.

Q: If the E' were actually .3V, instaed of .285V, what would be true of the pH measured by the pH meter, assuming no correction is made?

The answer says that the pH determined will be too low, because E' in calculation is too high.

First of all, isn't the E' used in calculation too low, because the wrong value is used? And also, if E'used is smaller than E'actual, wouldn't the second log term also be smaller, so H+ is lower, and pH is higher? I rewrote the equation above as Eobs= E' + .592pH, in which case the pH term should be bigger for smaller E'.

I'm not sure where I'm stuck.. please feel free to correct me where I'm wrong and any extra explanation would be very helpful. Thanks so much~

Use the equation they gave you to solve for pH

Eobs= E'cell- .0592*log(H+)

1. Eobs-E'cell= -0.592 x log[H+]

but since (pH= -log[H+]) we can rewrite this eequation as

2 .Eobs= E' + .592pH (I see you did this as well)

Continue to solve for pH by subtracting E' from both sides.

3. Eobs-E'=.592pH

Divide both side by .592
(Eobs-E')/.592 =pH

Now if we increase E' ( which is what we are doing since we are increasing it from .285V to .3V)
what happens to the numerator? Does it decrease or increase? What happens to the overall fraction if we assume Eobs is constant and increase E'?

If you said decrease then you are right!

the Actual pH will be lower than the measured

hahhaha i got the same answer as you. anyway where did you get this question from?

 

[RogueUnicorn]

I rewrote the equation above as Eobs= E' + .592pH, in which case the pH term should be bigger for smaller E'.

you're soooo close! but read the question again, the question says the E' is LARGER, so pH has to be lower. also, remember that the question states you're measuring Eobs, so that must stay constant.

edit: this is a kaplan question, right? it's not a very good one i must say.

 

[evusq]

yeah, I see that Eobs is constant. But for the actual value is larger, using the wrong lower value, wouldn't the measured pH be higher than the actual pH, which you get using the right value?

do you think this is a wrong answer? or am i mistaken? i undrestand both of your explanations which seem to point to the answer i got. (i picked that the measure pH would be higher, because E' in calculation is too low)

 

[RogueUnicorn]

the question is just bad; ignore it.. i know it's tough to swallow, but it's just not a fair q because it's ambiguous.