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frutty [35]
3 years ago
12

What is the (OH-) in a solution with a pOH of 6.48

Chemistry
1 answer:
inn [45]3 years ago
3 0
Through manipulation of equations, we are able to obtain the equation:

-pOH= log [ OH^{-}]

Then we can transform the equation into:

[ OH^{-}]= 10^{-pOH}

Then we are able to plug in the pOH and directly get [OH-]:

[ OH^{-}] = 10^{-6.48}

[ OH^{-}]=3.31* 10^{-7} M
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2. 2.74 L

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Generally speaking, you want to convert units to SI units, but in this case, we are working with ratios.  This makes up for using the units that wouldn't appropriate elsewhere.

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V₂ = 2.74 L.  

3.  Use the equation V₁/T₁ = V₂/T₂. Solve for T₂.  

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T₂ = 488 K.

4.  Do the same as above, but for V₂.  

(5.10 L)/(-56°C) = V₂/(-82°C)  

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5.  Use the equation V₁/n₁ = V₂/n₂.  Solve for V₂.  

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6.  Do the same as above, but for n₂.

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n₂ = 2.85 mol

7.  Use the equation P₁/T₁ = P₂/T₂.  Solve for T₂.  

(3.00 atm)/(390 K) = (2.45 atm)/(T₂)  

T₂ = 319 K

8.  Do the same as above, but for P₂.  

In this specific case, however you will need to convert units.  Since both temperatures don't have the same sign, the ratio won't come out right.  Convert to Kelvin.  Add 273.15 to the temperature in Celsius to convert to Kelvin -12.3°C = 260.85 K  25°C = 298.15 K.

(3.00 kPa)/(260.85 K) = P₂/(298.15 K)

P₂ = 3.43 kPa

There is a lot in here... If you are confused about something, let me know!

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