Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: JNW2 on March 18, 2016, 12:04:41 PM
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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.
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Calculate stoichiometry carefully.
Step 1: strong acid base react first then we have HCl left
This is not true statement.
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Step 1: strong acid base react first then we have HCl left
This is not true statement.
Why not?
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There are more moles HCl then KOH
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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?
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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.
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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?
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Your pH is 7.21, hence you buffer works on pKa2 of H3PO4.
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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?
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Work of buffers is based on common ion effect and such approximation is sufficient in majority of cases.
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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?
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You should solve a system of equations of type
x+y=c(phosphate)
y/x or x/y (from H-H equation)
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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.
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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?
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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 ?
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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
KOH reacts with HCl
excess of HCl 0.01 mole
HCl reacts with Na3PO4 => Na2HPO4
excess of HCl 0.005 mole
HCl reacts with KCN
excess of KCN 0.005 mole
HNO3 reacts with excess of KCN, then with some of Na2HPO4
From pH you should calculate how much HNO3 you need in this step.
This is quite simple stoichiometry if done stepwise.
This order comes from presence of strong acids and bases and pKas' order of week acids (the rule is: stronger acid replaces the weaker).
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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
KOH reacts with HCl
excess of HCl 0.01 mole
HCl reacts with Na3PO4 => Na2HPO4
excess of HCl 0.005 mole
HCl reacts with KCN
excess of KCN 0.005 mole
HNO3 reacts with excess of KCN, then with some of Na2HPO4
From pH you should calculate how much HNO3 you need in this step.
This is quite simple stoichiometry if done stepwise.
This order comes from presence of strong acids and bases and pKas' order of week acids (the rule is: stronger acid replaces the weaker).
The way you show is very simply and better than me but
why you know KCN will be formed?
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why you know KCN will be formed?
It won't, it is a typo.
Actually what happens (and matters) is
CN- + H+ :requil: HCN
Countercation (be it Na+ or K+) is completely irrelevant.
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@AWK. @Borek Thanks you very much
I conclude the basic concept about acid and base.
using concentration (ICE) first but if it is very complicated,use steps of Stoichemistry.
Right?
In HPO4-
If [H+ ]final =y and [H+ ]from HNO3 =x
we must approximate y<<<0.005- x right? to solve final equation ?
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Of course, NaCN (KCN predominantly is used in lab) and weak acid (instead of week).
The final H+ in buffer comes from equilibrium between H2PO4- and HPO4- and is used only in H-H equation.
In stoichiometric calculation for solution you assume the complete neutralization.
Show us (the final) volume of 0.100 M HNO3 you obtained. I do not know that you finished problem correctly.
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Of course, NaCN (KCN predominantly is used in lab) and weak acid (instead of week).
The final H+ in buffer comes from equilibrium between H2PO4- and HPO4- and is used only in H-H equation.
In stoichiometric calculation for solution you assume the complete neutralization.
Show us (the final) volume of 0.100 M HNO3 you obtained. I do not know that you finished problem correctly.
NaCN 0.0025 mole and Na2HPO4 0.0050 mole
if NaCN neutralize with HNO3 x mole ,there will be HCN x mole
and NaCN 0.0025-x mole then use buffer equation
:pH =pKa+ log HCN /NaCN
In the same volume ,thus concentration will change to mole.
Using this way for HPO4- and H2PO4, their conjugate acid . But use HNO3 y mole
Then the amount HNO3 used x+y mole and calculate its volume .
OK?
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Look at Kas and forget about NaCN i HCN! Such a buffer shows pH close to 10. You should neutralize NaCN completely and hydrogen phosphate partially.
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Look at Kas and forget about NaCN i HCN! Such a buffer shows pH close to 10. You should neutralize NaCN completely and hydrogen phosphate partially.
OK. You meant that Using HNO3 0.0025 mole to neutralize CN- ?
because Kb CN- = 1.6x10^(-5) >> Ka HCN = 6.2x10^(-10) ?
So, we ignore hydrolysis effect of HCN?
for HPO4 2- ,let x =amount of HNO3
Since pH =7.21 --> [H+]= 6.16x10^-8 and Ka H2PO4- = 6.2x10^-8
Thus, 6.2x10^-8 =6.16x10^-8 [HPO42-]/[H2PO4-]
1.0064= (0.005-x)/x
x = 2.492x10^-3 mole
all the needed amount of HNO3 = 4.992x10^-3 mole
So the final volume = 50 ml
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How many cm3 of 0.100 M HNO3 is needed for neutralization of
NaCN 0.0025 mole
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So, we ignore hydrolysis effect of HCN?
HCN does not hydrolyze, it eventually undergoes dissociation. This can be safely neglected in this case.
From H-H equation you may calculate ratio of [HPO42-]/[H2PO4-]
From your data you can calculate sum of concentrations of [HPO42-] and [H2PO4-]
Having know concentration of [H2PO4-] you can calculate the last portion of 0.100 M HNO3.
And note - you use Ka with 2 significant digits (this is important for the final volume of nitric acid).
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How many cm3 of 0.100 M HNO3 is needed for neutralization of NaCN 0.0025 mole
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So, we ignore hydrolysis effect of HCN?
HCN does not hydrolyze, it eventually undergoes dissociation. This can be safely neglected in this case.
From H-H equation you may calculate ratio of [HPO42-]/[H2PO4-]
From your data you can calculate sum of concentrations of [HPO42-] and [H2PO4-]
Having know concentration of [H2PO4-] you can calculate the last portion of 0.100 M HNO3.
And note - you use Ka with 2 significant digits (this is important for the final volume of nitric acid).
I have edited the solution and answer.
The answer is 50 mL.
Right?
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!
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!
You use only one exclamation mark. I have done it wrong?
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Almost correct.
Why?
50 is ambigous concerning significant digits - one or two digits (5·102 or 5.0·102).
50. is precise (no zero after dot).
Moreover, each problem after calculations, can be easily checked. In this one, even without calculator since during calculations you calculated pKa.
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@ AWK Thank you very much