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Topic: pH solutions in zumdahl chemistry books  (Read 6719 times)

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Offline AWK

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Re: pH solutions in zumdahl chemistry books
« Reply #15 on: March 21, 2016, 10:13:03 AM »
Quote
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).
AWK

Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #16 on: March 21, 2016, 10:21:05 AM »
Quote
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?

Offline Borek

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Re: pH solutions in zumdahl chemistry books
« Reply #17 on: March 21, 2016, 10:31:47 AM »
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.
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 #18 on: March 21, 2016, 10:59:22 AM »
@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 ?
« Last Edit: March 21, 2016, 12:44:42 PM by JSK2 »

Offline AWK

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Re: pH solutions in zumdahl chemistry books
« Reply #19 on: March 21, 2016, 01:48:24 PM »
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.
AWK

Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #20 on: March 21, 2016, 07:31:35 PM »
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?



Offline AWK

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Re: pH solutions in zumdahl chemistry books
« Reply #21 on: March 21, 2016, 08:06:55 PM »
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.
AWK

Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #22 on: March 22, 2016, 06:36:37 AM »
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


« Last Edit: March 22, 2016, 08:30:25 AM by JSK2 »

Offline AWK

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Re: pH solutions in zumdahl chemistry books
« Reply #23 on: March 22, 2016, 07:27:02 AM »
How many cm3 of 0.100 M HNO3 is needed for neutralization of
Quote
NaCN 0.0025 mole
?

Quote
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).
AWK

Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #24 on: March 22, 2016, 08:31:18 AM »
How many cm3 of 0.100 M HNO3 is needed for neutralization of
Quote
NaCN 0.0025 mole
?

Quote
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?

Offline AWK

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Re: pH solutions in zumdahl chemistry books
« Reply #25 on: March 22, 2016, 08:41:50 AM »
!
AWK

Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #26 on: March 22, 2016, 09:26:19 AM »
!
You use only one exclamation mark. I have done it wrong?

Offline AWK

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Re: pH solutions in zumdahl chemistry books
« Reply #27 on: March 22, 2016, 09:42:43 AM »
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.
AWK

Offline JNW2

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Re: pH solutions in zumdahl chemistry books
« Reply #28 on: March 22, 2016, 10:55:28 AM »
@ AWK Thank you very much

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