Calculate [H+] and [OH-] for each solution at
25°C
pH= 7.5
pH= 12.3
pH= 2.3
*please write in print (legibly)*
1) pH =7.5
pH=-log10[H+]
7.5=-log10[h+]
[H+] =antilog(-8.5)
[H+] =3.16*10^-8
PH+POh=14
POH=14-7.5=6.5
POH=-log10[OH-]
6.5=-log10[OH-]
[OH-]=antilog (-7.5)
[OH-]=3.16*10^-7
2)pH=12.3
pH=-log10[H+]
12.3=-log10[H+]
[H+]=antilog(-13.7)
[H+] =5.01*10^-13
pH+pOH=14
pOH=14-12.3=1.7
pOH=-log10[OH-]
1.7=-log10[OH-]
[OH-] =antilog (-2.3)
[OH-] =1.99*10^-2
3)ph=2.3
pH=-log10[H+]
2.3=-log10[H+]
[H+] =antilog (-3.7)
[H+] =5.01*10^-3
pH+pOH=14
pOH=14-2.3=11.7
pOH=-log10[OH-]
11.7=-log10[OH-]
[OH-]=antilog (-12.3)
[OH-]=1.99*10^-12
Calculate [H+] and [OH-] for each solution at 25°C pH= 7.5 pH= 12.3 pH= 2.3 *please...
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