00:02
Here in this problem given 5 .00 milliliter of 0 .011 molar acl is added to 50 milliliter of water.
00:13
We have to find the ph of the resulting solution.
00:17
Now for hydrochloric acid solution, we have the volume and we have the molar concentration.
00:25
So we can find out moles of acl that will be volume in later time.
00:34
Smaller concentration here volume given 5 milliliter 5 .0 milliliter will convert it into liter and 1 liter is 1000 milliliter so we multiply by 1 liter by 1 liter by 1 ,000 milliliter times smaller concentration this 1 .011 moles per liter we can sell little liter milliliter and we get 5 .5 times 10 to the power negative 5 moles.
01:14
Now we will find out total volume of the solution.
01:21
Total volume of the solution.
01:28
Initially the solution was 5 milliliter then 50 milliliter water was added.
01:36
So total volume 5 .00 plus 50 .00.
01:40
50 .00 milliliter.
01:44
That is 55 milliliter.
01:48
Now we'll convert it into liter, 55 milliliter times one liter divided by 1 ,000 milliliter because 1 liter means 1 ,000 milliliter.
02:00
We can sell milliliter, and we get 0 .055 liter.
02:07
Now we have number of more.
02:10
Moles of acl and we have the total volume of the solution in liter...