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Topic: pH solutions in zumdahl chemistry books  (Read 6883 times)

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Offline JNW2

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pH solutions in zumdahl chemistry books
« on: March 18, 2016, 12:04:41 PM »
Consider a solution prepared by mixing the following:
50.0 mL of 0.100 M Na3PO4
100.0 mL of 0.0500 M KOH
200.0 mL of 0.0750 M HCl
50.0 mL of 0.150 M NaCN
Determine the volume of 0.100 M HNO3 that must be added to this mixture to achieve a final pH value of 7.21.

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Attempt to solve problem:
Step 1: strong acid base react first then we have HCl left
Step 2: common ion effect
H+ by HCl dissociate completely will change equilibrium.
OH- by NaCN and Na3PO4 less, thus equilibrium change left to right.

Offline AWK

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Re: pH solutions in zumdahl chemistry books
« Reply #1 on: March 18, 2016, 01:58:49 PM »
Calculate stoichiometry carefully.

Quote
Step 1: strong acid base react first then we have HCl left
This is not true statement.
AWK

Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #2 on: March 18, 2016, 07:57:44 PM »


Quote
Step 1: strong acid base react first then we have HCl left
This is not true statement.
Why not?

Offline AWK

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Re: pH solutions in zumdahl chemistry books
« Reply #3 on: March 18, 2016, 09:59:12 PM »
There are more moles HCl then KOH
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Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #4 on: March 19, 2016, 02:31:01 AM »
There are more moles HCl then KOH
Ok, that's right. The remaining HCl will react with Na3PO4 before NaCN ? because PO43- stronger base than CN- . we will calculate hydrolysis of PO43-. But HCl will be left for NaCN .
But having NaCN more than HCl,there will be buffer solution.
Am I right?

 

Offline AWK

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Re: pH solutions in zumdahl chemistry books
« Reply #5 on: March 19, 2016, 02:56:04 AM »
At this moment pH is ~9.
Write down all pKa values for HCH (1) and for H3PO4 (3 values) an compare them with pH of buffer. Buffers work within ±1 pH unit  around pKa. Choose correct value of pka. Then calculate needed volume of HNO3. A few steps of calculations are still needed.
« Last Edit: March 19, 2016, 03:37:33 AM by AWK »
AWK

Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #6 on: March 19, 2016, 07:24:01 AM »
At this moment pH is ~9
after neutralize KOH , there is 0.010 mole HCl.
pKa. H3PO4 ~ 2.1 , pka2 ~ 7.2 ,pka3 ~11.5 ,pKa HCN ~ 9.3 
because pH must >  7.21 but not much basic. so, use pka HCN ,right?

« Last Edit: March 19, 2016, 09:14:39 AM by JSK2 »

Offline AWK

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Re: pH solutions in zumdahl chemistry books
« Reply #7 on: March 19, 2016, 03:50:33 PM »
Your pH is 7.21, hence you buffer works on pKa2 of H3PO4.
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Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #8 on: March 19, 2016, 08:34:52 PM »
Your pH is 7.21, hence you buffer works on pKa2 of H3PO4.
OK. But your approach ignore others acid hydrolysis .
Why this can be?

Offline AWK

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Re: pH solutions in zumdahl chemistry books
« Reply #9 on: March 19, 2016, 10:00:09 PM »
Work of buffers is based on common ion effect and such approximation is sufficient in majority of cases.
AWK

Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #10 on: March 19, 2016, 11:01:02 PM »
Work of buffers is based on common ion effect and such approximation is sufficient in majority of cases.
What can I do next? Do we have H2PO4- and HPO42- equal concentration? but I can't calculate it.
Then H+ from HNO3 change equilibrium . So, we use ICE table? 

Offline AWK

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Re: pH solutions in zumdahl chemistry books
« Reply #11 on: March 20, 2016, 03:56:10 AM »
You should solve a system of equations of type
x+y=c(phosphate)
y/x or x/y (from H-H equation)
AWK

Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #12 on: March 21, 2016, 04:09:54 AM »
You should solve a system of equations of type
x+y=c(phosphate)
y/x or x/y (from H-H equation)

More hints ,please .
I still can't solve this problem.

Offline Borek

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Re: pH solutions in zumdahl chemistry books
« Reply #13 on: March 21, 2016, 08:13:51 AM »
Assume KOH and HCl react first - that will just remove KOH and some HCl, simplifying the problem.

Then, list all pKa values involved (4 of them matter). At pH=7.21, which bases will be fully protonated, which will be not protonated at all, and which will be protonated partially? To what extent?

How much acid - in total - do you need to get to this situation? (Once you calculate amounts of bases protonated, this will be a simple stoichiometry). How much acid (HCl) do you have? ANd how much acid (HNO3) do you need to add?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #14 on: March 21, 2016, 09:30:56 AM »
Assume KOH and HCl react first - that will just remove KOH and some HCl, simplifying the problem.

Then, list all pKa values involved (4 of them matter). At pH=7.21, which bases will be fully protonated, which will be not protonated at all, and which will be protonated partially? To what extent?

How much acid - in total - do you need to get to this situation? (Once you calculate amounts of bases protonated, this will be a simple stoichiometry). How much acid (HCl) do you have? ANd how much acid (HNO3) do you need to add?
After neutralizing KOH, HCl remains 0.01 mole.
I think that PO43- (which the max Kb ) fully to be protonated.
So, that will be 0.0102 mole HPO42-and use all HCl to neutralize.
If I right, OH- remaining =0.0002 ----*
For HPO42- be protonated partially to form H2PO4-
Hence ,for pH =7.21 = pka +log base/acid =7.20+log([H2PO4-]/[HPO42- ]) ---------1)
CN- be protonated partially to form HCN?
Hence, pH =7.21= 9.3+log([HCN]/[CN]) --------2)
In this step, we will find the amount of H+ in ICE of these two ions ?
Then from *,1) ,2) we will get all amount H+ ion.
And we will get the amount of HNO3 ?

« Last Edit: March 21, 2016, 09:59:42 AM by JSK2 »

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