try

71 ka answer C hai

haan kaise??

yaar humein copy mein kuch Ka yad rakne ke liye kaha tha
uun mein se yeh ek hai :P

will try u the method after some time

Q71 .. H₂O -><- H⁺ + OH^-
[H₂O] = 55.5 moles/litre.
Now ionisation constant = K of equation = [H⁺][OH^-]/[H₂O] = K_w/55.5 = 10^-14/55.5 = 1.8*10^-16

72.

The solution clearly forms a buffer.

Using henderson equation -->

pOH = pK_b + log \frac{[salt]}{[base]}

[salt] = (0.01 * 100)/ 200
[acid] = (0.01 * 100)/200

pK_a + pK_b = 14 {for a particular conjugate acid-base pair}

pK_b = 14 - 9.26 = 4.74

pOH = 4.74 + log 1 = 4.74
Thus, answer is A.

Please correct me, if I'm wrong.
ALL THE BEST

i dunno i 2 gt d same answer bt woh answer ni hai

can u try d first q also