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Henry’s Law

Henry’s Law. S 1 S 2 = P 1 P 2. The question. If 0.80 g of sulfur dioxide at 10.00 atm pressure ( P1) dissolves in 5.00 L of water at 25.0°C, how much of it will dissolve in 1 L of water at 9.00 atm pressure ( P2) and the same temperature?. Pull out the important parts.

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Henry’s Law

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Presentation Transcript

  1. Henry’s Law

    S1S2 = P1P2

  2. The question

    If 0.80 g of sulfur dioxide at 10.00 atm pressure (P1) dissolves in 5.00 L of water at 25.0°C, how much of it will dissolve in 1 L of water at 9.00 atm pressure (P2) and the same temperature?

  3. Pull out the important parts

    0.80 g of sulfur dioxide, SO2 In 5.00 L At 25◦C At 10.00 atm, P1 X g In 1 L At 25◦C At 9.00 atm, P2

  4. Start assembling the equation

    First, Calculate S1 Solubility of a gas is measured in g/L So we have S1 = 0.80g 5.00 L = 0.16 g / L

  5. Solve for S2Another way of looking at the equation

    S1S2 = P1P2 Cross multiply to get an equation that is easy to use.

  6. Solve for S2Another way of looking at the equation

    S1S2 = P1P2 Cross multiply to get an equation that is easy to use. S1 P2 = S2 P1

  7. Solve for S2Another way of looking at the equation

    S1 P2 = S2 P1 We are solving for S2 so we’ll divide both sides by P1. Yes, we could have just multiplied both sides by P2. I am trying to think of ways to make this equation easy to remember- is it easier as S1P2 = S2P1 or as “S1 over P1 = S2 over P2”? Use what makes your brain happy. S1P2 P1 (0.16g/L )(9.00 atm) 10.00 atm = 0.144 g/L = S2

  8. It asked for how much will dissolve in 1 L.

    = x g 1 L 0.144 g L Hopefully the answer is obvious that 0.144 g will dissolve in 1 L. Just for fun, what if they asked how much would dissolve in 0.75L? = x g 0.75 L 0.144 g L Cross multiply to get (0.114 g) (0.75L) = (x g) (1 L) 0.0855 gL = x gL Divide both sides by L to get 0.0855 g will dissolve in 0.75 L
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