Solution.
"pH=-log[H_3O^+];" "[H_3O^+]=10^{-pH};"
If "pH=3.350;"
"[H_3O^+]=10^{-3.35}=4.467\\sdot10^{-4}M;"
If "pH=6.780;"
"[H_3O^+]=10^{-6.78}=1.66\\sdot10^{-7}M;"
"4.467\\sdot10^{-4}-1.66\\sdot10^{-7}=4467\\sdot10^{-7}-1.66\\sdot10^{-7}=4465.34\\sdot10^{-7}="
"=4.46534\\sdot10^{-4}M;"
Answer: concentration "[H_3O^+]" decreased by "4.46334\\sdot10^{-4}M."
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