Explain the voyage of water molecules as it boils in water
1 answer:
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Through manipulation of equations, we are able to obtain the equation:
![-pOH= log [ OH^{-}]](https://tex.z-dn.net/?f=-pOH%3D%20log%20%5B%20OH%5E%7B-%7D%5D%20)
Then we can transform the equation into:
![[ OH^{-}]= 10^{-pOH}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%3D%2010%5E%7B-pOH%7D%20%20)
Then we are able to plug in the pOH and directly get [OH-]:
![[ OH^{-}] = 10^{-6.48}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%20%3D%2010%5E%7B-6.48%7D%20)
![[ OH^{-}]=3.31* 10^{-7} M](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%3D3.31%2A%2010%5E%7B-7%7D%20M%20%20)
Then we can transform the equation into:
Then we are able to plug in the pOH and directly get [OH-]: