- what is the formula for degree of hydrolysis h, h= ( 10 to the power -7 (kakb)to the power -1/2) is applicable for which thing give some also exp for this

2- to a 200ml of 0.1 m weak acid ha soln 90ml of 0.1 m solnof naoh added what volm of 0.1 m naoh be added into above situation so that resulting ph is 5 ( ka(ha))=10 to the power -5

Please post your queries separately it becomes difficult for us to solve all of them simultaneously.

So i am answering your second query :

2) HA + NaOH  --------> Na⁺ + A^- + H₂O
   
No. of moles of  NaOH = 90 x 0.1 = 9 mili moles
No. of moles of HA = 200 x 0.1 =20 milimoles
No. of moles of  HA left  = 20-9 =11 milimoles
No. of moles of A- formed = 9 milimoles

Let the volume of NaOH added be  V
Now 
total no. of moles of A^- will be = 9+0.1V
pH=pka +log[salt/acid]
5= -log(10^-5) +log[(9+0.1V)/(11-0.1V)]

5=5 + log[(9+0.1V)/11-0.1V)]

Solving we get 
[(9+0.1V)/11-0.1V)] =1
9+0.1V =11-0.1 V
Thus
Volume of NaOH added  (V) =10 ml