At 25 °C, how many dissociated OH– ions are there in 1253 mL of an aqueous...

At 25 °C, how many dissociated OH– ions are there in 1253 mL of an aqueous solution whose pH is 1.97?

Expert Solution

PH = 1.97

POH = 14-PH

= 14-1.97

= 12.03

POH = 12.03

-log[OH-] = 12.03

[OH-] = 10^-12.03 = 9.33*10^-13M

no of moles of OH^- = molarity * volume in L

= 9.33*10^-13*1.253

= 1.17*10^-12moles

no of OH^- ions = no of moles of OH^- * 6.023*10^23

= 1.17*10^-12*6.023*10^23

= 7*10^11 ion of OH^- >>>>answer

queen_honey_blossom answered 2 years ago

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