Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 2 (Q.No. 3)
3.
A wastewater sample contains 10-5.6 mmol/l of OH ions at 25° C. The pH of this sample is
Discussion:
19 comments Page 2 of 2.
Bharath said:
8 years ago
@Sheela :
pH = -log10^-5.6-3 = -log10^-8.6 = 8.6.
then, 14-8.6 = 5.4.
pH = -log10^-5.6-3 = -log10^-8.6 = 8.6.
then, 14-8.6 = 5.4.
B.S.DALAL said:
7 years ago
Ph -log ^-5.6=-log10^-8.6 = (14-8.6) = 5.4.
Dhanu said:
7 years ago
How come 14 here?
Ibrahim said:
7 years ago
PH + POH = 14.
Since we are given a concentration of Oh ions we are calculating POH. To get PH we subtract from 14.
Since we are given a concentration of Oh ions we are calculating POH. To get PH we subtract from 14.
Dhanu said:
7 years ago
Thank you all for explaining this.
Suhail said:
5 years ago
Thanks all for explaining it.
Manjula said:
4 years ago
H+ * OH- = 10^-14.
H+ *(10^-5.6 *10^-3) = 10^-14.
H+= (10^-14)/(10^-8.6)
= 10^-5.4.
H+ *(10^-5.6 *10^-3) = 10^-14.
H+= (10^-14)/(10^-8.6)
= 10^-5.4.
(1)
Shashank pandit said:
3 years ago
poh = 8.6.
As the mmol will change to mole per lit;
therefore as we know;
pH + pOH = 14.
pH + 8.6 = 14.
pH = 14 - 8.6 = 5.4.
As the mmol will change to mole per lit;
therefore as we know;
pH + pOH = 14.
pH + 8.6 = 14.
pH = 14 - 8.6 = 5.4.
(2)
Swathy said:
8 months ago
How come 3 there, ie; 5.6-3?
Anyone, please explain this step.
Anyone, please explain this step.
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