00:05
Consider the following reaction along with the equilibrium expression for this reaction which is shown.
00:12
It turns out the equilibrium expression gives us the value of ka equivalent to 1 .8 into 10 ratio per minus 5.
00:22
We can calculate the moles of nh4cl, that is, 0 .102 moles.
00:29
The mole of os negative ion equals to 0 .0025.
00:35
Hence, the initial number of mole is also calculated as shown.
00:43
With the help of above data, initial concentration can be calculated using the same formula.
00:50
The initial and equilibrium concentration for this system is summarized into the tabular form over here.
00:59
Now substitute the equation, now substitute the equilibrium concentration into equilibrium expression and 1 .8 into 10 ratio power minus 5 for ka.
01:11
Assume that 0 .250 minus s is equivalent to 0 .250 and 0 .7502 minus x gives us 0 .7500 minus x gives us 0 .7500 minus x gives us 0 .5 .000.
01:29
0 .7520 substituting all the assumed values in the equation for kb which it gives us the value of x equals 6 .0 into 10 ratio by minus 6.
01:42
At equilibrium the concentration of hydroxide ion is calculated which is 6 .0 into 10 ratio minus 6 the value of p o h is negative value of concentration of o h negative ion it gives us 5 .22 we can calculate the total ph of solution by adding the ph and poh values.
02:09
It gives us 8 .78 for part b...