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Christineg wrote:After your calculation, how would you determine which value is your pH and your pOH?
For example in 6B. 5D how would you determine that 3.15 is your pOH and 10.85 is your pH?
Remember that the pH added to the pOH equals 14. This is because the sum of the OH- ion and H+ ion concentration must equal the equilibrium constant for the ionization of water (1x10^-14). If you take the -log(1x10^-14)=14. In short, the pH added to the pOH is equal to 14. Since 14-3.15=10.85, the pOH is 10.85. I hope this helps.
You should be able to look at the actual molecule to determine if it's an acid or a base. If it is an acid, you know that you are solving for the pH with the -log. If it is a base, you know that you are solving for the pOH with the log. Determining the opposite would just have pH + pOH = 14 as a total.
For 6B.5 part d, I first found M2 using M1V1=M2V2, and set M2 as my concentration of OH- ions. Then, I used -log([OH-]) to find pOH, which should give you 3.15. You would then subtract that from 14 because pH+pOH=14, and you should get 10.85.
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