Answer :
Answer: 1. pH= 8.0, basic
2. pH= 2.2, acidic
3. pH=12.1, basic
Explanation:
pH or pOH is the measure of acidity or alkalinity of a solution.
pH is calculated by taking negative logarithm of hydrogen ion concentration.
Acids have pH ranging from 1 to 6.9, bases have pH ranging from 7.1 to 14 and neutral solutions have pH equal to 7.
1. [tex][H_3O^+]=9.5\times 10^{-9}[/tex]
[tex]pH=-\log [H_3O^+][/tex]
Putting in the values:
[tex]pH=-\log[H_3O^+]
[tex]pH=-\log[9.5\times 10^{-9}]=8.0/tex]
Solution is basic.
2. [tex][H_3O^+]=6.1\times 10^{-3}[/tex]
[tex]pH=-\log [H_3O^+][/tex]
Putting in the values:
[tex]pH=-\log[6.1\times 10^{-3}]=2.2/tex]
Solution is acidic.
3. [tex][OH^-]=1.3\times 10^{-2}[/tex]
[tex]pOH=-\log [OH^-][/tex]
Putting in the values:
[tex]pOH=-\log[1.3\times 10^{-2}]=1.9[/tex]
pH+pOH=14
pH=14-1.9=12.1
Solution is basic
Following are the solution to the PH values:
Given:
[tex]\to [H_3O^+ ] = 9.5 \times 10^{-9}\ M \\\\[/tex]
[tex]\to [H_3O^+ ] = 6.1 \times 10^{-3}\ M \\\\[/tex]
[tex]\to [OH^- ] = 1.3 \times 10^{-2}\ M[/tex]
To find:
base / acidic=?
Solution:
For point 1:
[tex]\to PH= -\log [H_3 O^{+}]\\\\[/tex]
[tex]=-\log(9.5 \times 10^{-9})\\\\=8.022 \\\\[/tex]
Therefore [tex]PH> 7 \to[/tex] it is base.
For point 2:
[tex]\to POH= -\log [H_3 O^{+}]\\\\[/tex]
[tex]=-\log(6.1 \times 10^{-3})\\\\=2.214 \\\\[/tex]
Therefore [tex]PH< 7 \to[/tex] it is acid.
For point 3:
[tex]\to PH= -\log [OH^{-}]\\\\[/tex]
[tex]=-\log(1.3 \times 10^{-2})\\\\=2- \log 1.3 \\\\=2 -0.1139433523\\\\= 1.88605664769\\\\[/tex]
[tex]\therefore[/tex]
[tex]\to PH+POH=14\\\\\to PH=14 - POH\\[/tex]
[tex]= 14- 1.88605664769\\\\= 12.1139433523\\\\\therefore,[/tex]
[tex]PH> 7 \to[/tex] it is base.
Therefore, the answer is "base, acid, and base".
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