Answers

2014-11-24T02:11:41+05:30

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
I understand the question this way.  I hope it is correct.
Into and from a container, there are incoming and outgoing water flows at the same rate V litres per hour.   Let us say there are always V2 litres in the container.

concentration of [ H+] in incoming water = 10^{-7.2} moles / litre
concentration of [ H+] ions in outgoing water = 10^{-8.4} moles / litre

So number of ions which came in t hours  = 10^{-7.2} V t   moles
Number of ions which went out = 10^{-8.4} V t   moles

So concentration of [ H+ ]ions in the container at time t hours = 
      { 10^{-7.2} - 10^{-8.4} ) V t  = V t * [ (10^{1.2} - 1) ] * 10^{-8.4}   

Average number of [ H+ ] in moles per litre at t hours : 
        =  V t * [ (10^{1.2} - 1) ] * 10^{-8.4} / V2
 
      [ H+ ]   = [ (10^{1.2} - 1) ] * 10^{-8.4} * (V t / V2)  moles

Taking  logs :     pH  = -7.228 + Log_10 (V t / V2)

The average pH value in the container increases continuously as the time progresses.  We are letting in more number of H+ ions and draining only a fraction of that.

0