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Topic: pH of sodium hydroxide + water  (Read 16961 times)

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Offline HmmHi

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pH of sodium hydroxide + water
« on: March 13, 2011, 12:41:53 PM »
I have 1 mL of 0.2M sodium hydroxide being added to 30 mL of water

NaOH(aq) + H2O(l)   :rarrow:  H2O(l)  +  NaOH(aq)

Which means  ???
-log [NaOH] should = pOH?
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Offline stewie griffin

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Re: pH of sodium hydroxide + water
« Reply #1 on: March 13, 2011, 02:23:22 PM »
p always stands for -log
So pOH is -log[OH-].
You need to determine if NaOH is a strong or weak base (will it fully dissociate?) before you can find the concentration of hydroxide ion.
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Offline HmmHi

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Re: pH of sodium hydroxide + water
« Reply #2 on: March 13, 2011, 03:32:50 PM »
Sodium hydroxide is a strong base, it dissociates completely

NaOH(aq)   :rarrow:  Na+(aq) + OH-(aq)
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Offline rabolisk

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Re: pH of sodium hydroxide + water
« Reply #3 on: March 13, 2011, 04:00:26 PM »
Ok. So what happens if you add more water to an existing solution of OH-?
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Offline HmmHi

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Re: pH of sodium hydroxide + water
« Reply #4 on: March 13, 2011, 04:22:43 PM »
You dilute it?  ???
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Offline rabolisk

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Re: pH of sodium hydroxide + water
« Reply #5 on: March 13, 2011, 04:53:57 PM »
Ok. So figure out the concentration after dilution.
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Offline HmmHi

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Re: pH of sodium hydroxide + water
« Reply #6 on: March 13, 2011, 05:25:56 PM »
Um, so

civi  =  cfvf
0.2M x 1mL  =  cf x 31mL
cf = 0.2M/31mL
= 0.0064516129M
-log[0.0064516129M] = 2.19
Like that  ???
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Offline rabolisk

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Re: pH of sodium hydroxide + water
« Reply #7 on: March 13, 2011, 05:48:48 PM »
Yes.
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Offline stewie griffin

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Re: pH of sodium hydroxide + water
« Reply #8 on: March 14, 2011, 09:24:00 AM »
Also don't forget that you just calculated a pOH, not a pH. It wouldn't make sense to have a pH of about 2 for a solution of NaOH.
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