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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: HmmHi on March 13, 2011, 12:41:53 PM

Title: pH of sodium hydroxide + water
Post by: HmmHi 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?

Title: Re: pH of sodium hydroxide + water
Post by: stewie griffin 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.

Title: Re: pH of sodium hydroxide + water
Post by: HmmHi on March 13, 2011, 03:32:50 PM

Sodium hydroxide is a strong base, it dissociates completely

NaOH_(aq)   :rarrow:  Na⁺_(aq) + OH^-_(aq)

Title: Re: pH of sodium hydroxide + water
Post by: rabolisk on March 13, 2011, 04:00:26 PM

Ok. So what happens if you add more water to an existing solution of OH^-?

Title: Re: pH of sodium hydroxide + water
Post by: HmmHi on March 13, 2011, 04:22:43 PM

You dilute it?  ???

Title: Re: pH of sodium hydroxide + water
Post by: rabolisk on March 13, 2011, 04:53:57 PM

Ok. So figure out the concentration after dilution.

Title: Re: pH of sodium hydroxide + water
Post by: HmmHi on March 13, 2011, 05:25:56 PM

Um, so

c_iv_i  =  c_fv_f
0.2M x 1mL  =  c_f x 31mL
c_f = 0.2M/31mL
= 0.0064516129M
-log[0.0064516129M] = 2.19
Like that  ???

Title: Re: pH of sodium hydroxide + water
Post by: rabolisk on March 13, 2011, 05:48:48 PM

Yes.

Title: Re: pH of sodium hydroxide + water
Post by: stewie griffin 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.