[AWK]

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 pK_as' order of week acids (the rule is: stronger acid replaces the weaker).

[JNW2]

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 pK_as' 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?

[Borek]

why you know KCN will be formed?

It won't, it is a typo.

Actually what happens (and matters) is

CN^- + H⁺ HCN

Countercation (be it Na⁺ or K⁺) is completely irrelevant.

[JNW2]

@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 ?

[AWK]

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.

[JNW2]

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?

[AWK]

Look at K_as and forget about NaCN i HCN! Such a buffer shows pH close to 10. You should neutralize NaCN completely and hydrogen phosphate partially.

[JNW2]

Look at K_as 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

[AWK]

How many cm³ of 0.100 M HNO₃ is needed for neutralization of

NaCN 0.0025 mole

?

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 [HPO₄^2-]/[H₂PO₄^-]
From your data you can calculate sum of concentrations of [HPO₄^2-] 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).

[JNW2]

How many cm³ of 0.100 M HNO₃ is needed for neutralization of

NaCN 0.0025 mole

?

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 [HPO₄^2-]/[H₂PO₄^-]
From your data you can calculate sum of concentrations of [HPO₄^2-] 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?

[AWK]

!

[JNW2]

!

You use only one exclamation mark. I have done it wrong?

[AWK]

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.

[JNW2]

@ AWK Thank you very much