00:02
Here in this problem, given concentration of nh4 br is 0 .55 molar and kb for nhc is given here.
00:12
We have to find the ph of the ammonium bromide solution.
00:19
Now, nh4 br, let's write the equation of the dissociation, produces nh4 plus and br minus.
00:34
Now, nh4 plus is the conjugate acid of weak base ammonia.
00:40
So it will react with water.
00:43
But br minus, that's the conjugate base for strong acid hbr.
00:51
So it will not react with water.
00:54
But nh4 plus will react with water as it is the conjugate acid of weak base ammonia.
01:01
So now first find out we have the.
01:05
Kb of n h3 so we'll find out k a for n h 4 plus now we know for a conjugate acid base pair k a times kb is equal to k w k w is the iron product constant of water so k a for n h four plus will be k w divided by kb for ammonia now k w value is 1 .0 times 10 to the power negative 14 and kb for n h3 given here 1 .76 times 10 to the power negative 5 and we get k a for n h 4 plus ion is 5 .68 times 10 to the power negative 10 now we'll write the equation for the dissociation of nh4 plus in aqua solution plus water produces nh3 and hydrogen ion.
02:19
Now let's set up ice table.
02:22
Initial concentration of nh4 plus is 0 .55 molar.
02:28
So let's write that 0 .55 molar here initial construction 0 and 0 .0 .0 .0 .0.
02:36
Concentration say minus x here positive x positive x so equilibrium concentration will be initial concentration minus x and here 0 plus x also x here now we'll write the expression for k a that will be concentration of n h3 times concentration of h3 o plus divided by equilibrium concentration of n h4 plus now so we'll write k a is concentration of n h3 x hydurneum ion x so x squared divided by equilibrium concentration of n h4 plus and that one is 0 .55 minus x and here we'll write the k a value this one 5 .68 times 10 to the power negative 10 x squared divided by here we see the value for k a is very small it is in the range of 10 to the power negative 10 so we assume the value for x is much much smaller than initial concentration 0...
