given
[OH-]= 7.2 x10^-2 M
pOH - -log[OH-] =-log[7.2 x10^-2 M]
= 3.14
***********
pH +POH = 14
so pH = 14- 3.14
= 10.86
**************
2)
given
pOH = 3.9
pH = 14-3.9 = 10.1
[H3O+] = 10^-pH = 10^-10.1
= 8 x10^-11 M
*******************
3)
a)
given
pH = 3
[H3O+] = 10^-pH = 10^-3
= 1x10^-3 M
******************
b)
pH of water = 7
so [H3O+] = 10^-7 M
ratio of [H3O+]cola : [H3O=]water = 10^-3/10^-7 = 10^4
Thus [H3O+]cola is 10000 times greater
****************************************
4)
given
pH = 3
[H3O+] = 10^-3 mols/L
so moles of H3O+ in 0.1 L = 10^-3 * 10^-1 = 10^-4
thus answer is
1 x10^-4 mols
*****************
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