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`        Calc. the pH of 10^-8 M acid sol.`
6 years ago

SAGAR SINGH - IIT DELHI
879 Points
```										Dear student,
Find the dissociation constant first and then write log[H+]
pH=-log[H+]

All the best.
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Sagar Singh
B.Tech, IIT Delhi

```
6 years ago
510 Points
```										10-8M acid = 10-8M H+ ion concentration...
in solution H+ from water is 10-7 M
total H+ = 10-7+10-8 = 1.01*10-7
pH = -log[H+] =-log[1.01*10-7]
= 6.85       approx
```
6 years ago
Sudheesh Singanamalla
114 Points
```										Molarity = 10^-8 .
n/V = 10^-8
n = 10^-8 * V
pH = -log [H+] .
Water normally dissociates:  H2O <===> H+  + OH-  The dissociation constant is Kw = 10^-7
So you would have 10^-7 moles of H+ in pure water.  Add the 10^-8 moles  of acid, and the resulting concentration of H+ is [H+] = 1.1 x 10^-7  the negative log of 1.1 x 10^-7 = 6.96, which is the pH value.
therefore pH of the solution is 6.96
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```
6 years ago
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