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Topic: Molar solubility and Buffer soultion  (Read 5174 times)

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Offline Grandmama

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Molar solubility and Buffer soultion
« on: December 14, 2007, 03:27:45 PM »
Hello, I would just like to ask a couple of questions. I have tried the problems out for myself and haven't been able to come up with a feasible answer.


(a)calculate the molar soulubility of silver carbonate in 0.1000M ammonium hydroxide
This problem I formulated a complex and after doing the steps came up with 7.6*10^-15


(b)how many mL of 0.1000M sodium hydroxide must be added to 500.0 mL of 0.1000M formic acid to prepare a buffer solution with a pH of 4.00?

This problem I used henderson hasselbeck and came up with 850mL

(1)what will be the resulting pH if 1.000 mL of 1.000M HCL is added to 100.0 mL of the buffer prepared in (b)?

Haven't tried, didn't know if buffer solution was right
(2)What will be the resulting pH if 2.000 mL of 1.000M NaOH is added to 200.0 mL of the buffer solution prepared in (b)?
Haven't tried, didn't know if buffer solution was right

Offline Borek

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Re: Molar solubility and Buffer soultion
« Reply #1 on: December 14, 2007, 05:43:32 PM »

(b)how many mL of 0.1000M sodium hydroxide must be added to 500.0 mL of 0.1000M formic acid to prepare a buffer solution with a pH of 4.00?

This problem I used henderson hasselbeck and came up with 850mL

You did something wrong - which is visible at the first sight (if you know what to look for ;) ). Both solutions (NaOH and formic acid) have the same concentration, so to compare number of moles of acid and base it is enough to compare volumes. Buffer solution must have some not neutralized acid left, yet you are adding excess NaOH (850 mL vs 500 mL). pH of this solution is around 12.4.

Could be 850 is the final volume of your buffer?
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Offline Grandmama

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Re: Molar solubility and Buffer soultion
« Reply #2 on: December 14, 2007, 07:01:56 PM »

(b)how many mL of 0.1000M sodium hydroxide must be added to 500.0 mL of 0.1000M formic acid to prepare a buffer solution with a pH of 4.00?

This problem I used henderson hasselbeck and came up with 850mL

You did something wrong - which is visible at the first sight (if you know what to look for ;) ). Both solutions (NaOH and formic acid) have the same concentration, so to compare number of moles of acid and base it is enough to compare volumes. Buffer solution must have some not neutralized acid left, yet you are adding excess NaOH (850 mL vs 500 mL). pH of this solution is around 12.4.

Could be 850 is the final volume of your buffer?

I calculated the volume on NaOh to be 310ml I used the Henderson Hasselbeck equation the wrong way. This time formic acid and NaOh give me the conjugate base which is then divided by the remaining acid concentration. Once you do that you take the log of that number and add it to the pka which game me 3.98216 using only three sig figs in my orginal answer I cam up with 310 mL and  500mL of formic acid for a total of 810mL.

Thanks

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