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posted by  FRAPPY on 3/17/2008 1:20:34 PM  |  status: Live  

quick, easy question, need asap, will rate life saver

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0.90 grams of sodium hydroxide (\rm NaOH) is dissolved in water to make 2.5 L of solution. What is the pH of this solution?  

I know the pH of the solution is 11.95.  How do I calculate the pOH of the solution?  Please give brief explanation.  thank you!
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posted by jks777 on 3/17/2008 1:30:24 PM  |  status: Live
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FRAPPY's comment:
"thank you!"
Response Details:
To find the pH of the situation, you will need to first find the concentration of OH- in the solution by using M = n/V and n = m/MM, where M is molarity/concentration, n is moles, V is volume in liters, m is mass and MM is molar mass.  In sodium hydroxide, the molar mass = 22.99 + 16 + 1.01 = 40 g/mol, therefore.
 
n = 0.9 g/40 g/mol = 0.0225 mol
 
And then molarity can be calculated.
 
M = n/V = 0.0225 mol / 2.5 L = 0.009 M
 
Then the pOH can be found, by the equation: pOH = -log[OH], where [OH] is the concentration of the solution.
 
pOH = -log(0.009)
pOH = 2.05
 
To find the pH, the pH + pOH must be equal to 14, thus:
 
pH + 2.05 = 14
pH = 11.95
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