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Sagot :
REmember that pH is the negatige logarithem of the concentration of protons:
[tex]pH=-\log _{10}\lbrack H^+\rbrack[/tex]when we make the invers of the logarithm:
[tex]\lbrack H^+\rbrack=10^{-pH}[/tex]in this case:
[tex]\lbrack H^+\rbrack=10^{-5}[/tex]if the concentration of protons undergoes a 1000-fold increase the new concentration is:
[tex]\lbrack H^+\rbrack=10^{-5}\times10^3=10^{-2}[/tex]therefore the new pH is calculated as follows:
[tex]pH=-\log _{10}\lbrack H^+\rbrack=-\log (10^{-2})=2[/tex]B.
For the second part we need to remember that [H+] and [OH-] are related according the following equation:
[tex]\lbrack H^+\rbrack\times\lbrack OH^-\rbrack=10^{-14}[/tex]We have calculated before that
[tex]\lbrack H^+\rbrack=10^{-5}[/tex]Then we can calculate [OH-]:
[tex]10^{-5}\times\lbrack OH^-\rbrack=10^{-14}\text{ }\rightarrow\lbrack OH^-\rbrack=10^{-9}\text{ }[/tex]1000 fold that concentration is
[tex]\lbrack OH^-\rbrack=10^{-9}\times10^3=10^{-6}[/tex]Again we use the relation between [H+] and [OH-] with the new value:
[tex]\lbrack H^+\rbrack\times\lbrack OH^-\rbrack=10^{-14}[/tex][tex]\lbrack H^+\rbrack\times10^{-6}=10^{-14}[/tex][tex]\lbrack H^+\rbrack=10^{-8}[/tex]And once again the pH formula:
[tex]pH=-\log _{10}\lbrack H^+\rbrack=-\log (10^{-8})=8[/tex]
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