[JNW2]

Consider solution formed by mixing 50 ml of 0.100 M H2SO4 , 30ml of 0.100 M HOCl,25 ml of 0.200 NaOH,25 ml of 0.100 M Ba(OH)2, and 10.0 ml of 0.150 M KOH. Find pH of this solution
..
I think, we must first neutralize H2SO4 and NaOH then KOH and Ba(OH)2 respectively but I don't sure that when HOCl neutralize.

Sorry for my poor english

JSK2

[Borek]

Assume strong acids get neutralized first.

[AWK]

Show, how you do it. Note, BaSO₄ is insoluble.

[JNW2]

Step 1: 0.1 M 50 ml H2SO4 and 0.2 M 25 ml NaOH
there is [HSO4-] left 0.1x50/140 M ,right?

Step 2: then HSO4- and 0.15 M 10 ml KOH
There are [SO42-] ,[HSO4-] left ,right?
Use Handelson hassenbach?

Step 3: 0.1 M 25 ml Ba(OH)2  will Ba(OH)2 left?  and
that BaSO4 not soluble , so there isn't any so4 2- left ?

Step 4: 0.1 M 25 ml HOCl and Ba(OH)2

I calculate it roughy . Am I right?

[AWK]

In the first step neutralize Ba(OH)₂ with H₂SO₄ (which is in excess to Ba(OH)₂).
Then neutralize the rest of sulfuric acid.
Finally neutralize the weak acid.
During calculation pH you should take into account a final volume.

Always write down balanced reactions.

[JNW2]

Why Ba(OH)2 neutralize first?
Not  KOH or NaOH ?

Isn't that involving base strength ?

So,consider delta  EN. more electronegative alkali ,less electronegative in oxygen ,weaker base. Am I right?
So, Ba(OH)2 > KOH >NaOH right?

[AWK]

Ba(OH)₂ is also a strong base. Moreover it forms insoluble barium sulfate.

[Borek]

Order in which you neutralize strong bases doesn't matter much. Actually you can't say which base reacts with the acid, as they are completely dissociated and cations involved - Na⁺, K⁺ and Ba²⁺ - are just spectators.

The precipitation doesn't matter much for the final pH, but it is definitely easier to take care of the Ba(OH)₂+H₂SO₄ reaction first.

You will need K_a of HOCl though (which also means the final volume should not matter much).

[AWK]

The final volume does matter in this case.

[JNW2]

HSO4- 0.1x150/140 M

[AWK]

Multiplying volume in cm³ by concentration in mole/dm³ (M) you calculate all amounts of reagents in milimoles. This is completely correct approach. But this should be mentioned correctly.
Stoichiometry of solution is calculated in moles or milimoles according to balanced reaction.
Looking at your numbers at first look I found the shortest way of calculation:
1. barium salt with sulfuric acid
2. the rest of sulfuric acid with NaOH
3. HOCl with KOH.
Write down these reactions and calculate stoichiometry.
All other approaches need more reactions and calculations.
Forget about Ksp. Just think you remove barium completely because  the low solubility of BaSO₄.

Alternatively you may calculate concentrations of all reagents in the final solution of 140 cm³. Then you may do stoichiometry according to balanced reactions just by addition or subtraction of concentrations, respectively. But in this case be careful with rounded numbers.

After finishing stoichiometry decide which formula you need for pH calculations.

Show us your calculation according both methods.

According to Borek suggestion you may treat all metal cation as spectator ions.
Then you can calculate only reactions of H⁺ with OH^-. But this approach is a bit more advanced.

[JNW2]

@AWK and @ Borek
Thank you very much.
I think I understand it now.
Step 1 and 2 balance pH because there is SO42- pKb very less
Step 3 HOCl and KOH there will be HOCl left with its salt
Thus, this is buffer solution. Using Handelson hassenbach ,
pH=pKa +log [conjugate base]/[acid].
= 7.53+log 10/20 = 7.53-0.3 =7.50

Am I right?

[AWK]

pH=pKa +log [conjugate base]/[acid].
= 7.53+log 10/20 = 7.53-0.3 =7.50

Am I right?

No, you are not right.
10 and 20  are not correct numbers.

[JNW2]

Sorry for my mistake
pH =7.53 because [OCl ]=[HOCl]
right?

[AWK]

At last. But this work only if you neglect ionic strenth (this is more advanced chemistry). This is why I suggested calculation of concentrations since for ionic strenth concentrations are absolutely needed, when Borek suggested the shortcut way without concentrations. Advanced students should work according to my way of solution.