00:01
Hi, in this question we have given this buffer system ch3 nh2 and ch3 nh3 nh3 plus cl minus.
00:06
If 5 .38 milliliters 0 .10 molar h cell is added, what will be ph? from henderson equation, we can write poh is equal with pkb plus log of concentration of salt, that is, ch3, nh3 plus, cl minus by concentration of ch3 in h2.
00:28
We will consider that volume of the buffer will remain constant that is be equal to constant now ch3 nh2 will react with h plus and cl minus to form ch3 nh3 plus cl minus number of moles of ch3 nh2 will decrease as h plus is reacting with this now number of most of hcl is 5 .38 into 0.
00:59
0 .10 equal to 0 .538 mill.
01:05
C .h3nh2 number of moles is 52 .045 millimole and number of moles for ch3 nh3 plus cl minus will be after multiplying this 171 .03 millimule.
01:24
So final number of moles of ch3 nh2 will be 52 .045 minus 0 .0.
01:35
538 equal to 51 .507 mole or millimole...