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If the ph of incoming and outgoing waters are 7.2 and 8.4 average ph value of water

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by vamsicivil34

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by vamsicivil34

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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.

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 :