Plz solve it.
Dear student,
First we need to find the concentration of H+ in the above solutions.
We know , pH = -log[H+] based on this we get,
Solution A:
Given pH = 2 ,
2 =-log[H+] , [H+] = 10-2 M
Solution B :
Given pH = 9 ,
9 =-log[H+] , [H+] = 10-9 M
Now we need to calculate the resultant molarity when these two are mixed.
MV = M1V1 + M2V2
Here V = 1000 mL , V1 = 100 mL and V2 = 400 mL
Thus substituting the values we get,
M x 1000 = 10-2 x 100 + 10-9 x 400
M x 1000 = 1 + 0.0000004 = 1.0000004 ( ~ 1 )
M = 1 / 1000 = 1 x 10-3
Thus resultant concentration will be 1 x 10-3 M .
Thus now pH = -log[H+] = -log[1 x 10-3 ] = 3-0= 3 .
Thus the resultant pH will be 3 .
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards
First we need to find the concentration of H+ in the above solutions.
We know , pH = -log[H+] based on this we get,
Solution A:
Given pH = 2 ,
2 =-log[H+] , [H+] = 10-2 M
Solution B :
Given pH = 9 ,
9 =-log[H+] , [H+] = 10-9 M
Now we need to calculate the resultant molarity when these two are mixed.
MV = M1V1 + M2V2
Here V = 1000 mL , V1 = 100 mL and V2 = 400 mL
Thus substituting the values we get,
M x 1000 = 10-2 x 100 + 10-9 x 400
M x 1000 = 1 + 0.0000004 = 1.0000004 ( ~ 1 )
M = 1 / 1000 = 1 x 10-3
Thus resultant concentration will be 1 x 10-3 M .
Thus now pH = -log[H+] = -log[1 x 10-3 ] = 3-0= 3 .
Thus the resultant pH will be 3 .
Hope this information will clear your doubts about topic.
If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.
Regards