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pH question finding the change

Screenshot_20161028-232752~2.pngi have tried to solve these questions but I still was not able to obtain the right answer. Where have I gone wrong ?
Thanks.
(edited 7 years ago)
Reply 1
14776941157261292128075.jpg here is a better picture .
Original post by coconut64
14776941157261292128075.jpg here is a better picture .


question 6 you have written the initial [HCl] wrongly...

OK I see what you have done.

You are calculating pH from the number of moles NOT from the concentration.

pH = -log[H+]
(edited 7 years ago)
Reply 3
Original post by charco
question 6 you have written the initial [HCl] wrongly...

OK I see what you have done.

You are calculating pH from the number of moles NOT from the concentration.

pH = -log[H+]


Appreciate your help. I think I managed to get it at the end. it is such a silly mistake and I will make sure I do proper working next time.1477760422264336308643.jpg
Reply 4
Original post by charco
question 6 you have written the initial [HCl] wrongly...

OK I see what you have done.

You are calculating pH from the number of moles NOT from the concentration.

pH = -log[H+]


For question7, I obtained 11.3 ( 13.176-1.903) for the pH change. However the answer is 11.58. Have I got it wrong?
Original post by coconut64
For question7, I obtained 11.3 ( 13.176-1.903) for the pH change. However the answer is 11.58. Have I got it wrong?


I got 13.17609126 - 1.602059991 = 11.574

If you round to two dp for each stage of the calc. you get

13.18 - 1.60 = 11.58
Reply 6
Original post by TeachChemistry
I got 13.17609126 - 1.602059991 = 11.574

If you round to two dp for each stage of the calc. you get

13.18 - 1.60 = 11.58


How did you get 1.60? For the conc of excess H+ ions I got 0.0125...

Thanks
Original post by coconut64
How did you get 1.60? For the conc of excess H+ ions I got 0.0125...

Thanks


You are correct. My mistake.
Reply 8
Original post by TeachChemistry
You are correct. My mistake.


are you sure because that's the answer...

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