What would be the pH, the hyronium ion concentration, and the hydroxide ion concentration in a solution prepared by dissolving 10.6 grams of solid NaOH in enough water to make 0.500L of solution?
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pH is equal to the negative log of the concentration of H+ ions. pOH is the negative log of the concentration of OH- ions. pH+pOH=14
For this problem there is one OH- for every molecule, and as such
NaOH --> 1 OH- + 1Na+
Then you multiply this value (1) by the Molarity, or mols/L of your solution.
In this case you have 10.6g and you divide this by the Molecular weight(40) to get .265mol NaOH
Next, you divide this by the liters in your solution .265/.5=.53M
And finally, you take the negative log of the Molarity to get your pOH. Which then 14-pOH=pH
The answer is 12.72
For this problem there is one OH- for every molecule, and as such
NaOH --> 1 OH- + 1Na+
Then you multiply this value (1) by the Molarity, or mols/L of your solution.
In this case you have 10.6g and you divide this by the Molecular weight(40) to get .265mol NaOH
Next, you divide this by the liters in your solution .265/.5=.53M
And finally, you take the negative log of the Molarity to get your pOH. Which then 14-pOH=pH
The answer is 12.72