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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Chemistry => Topic started by: ice_blockie on November 12, 2008, 04:15:03 pm

Title: pH question...
Post by: ice_blockie on November 12, 2008, 04:15:03 pm: from STAV exam:

A student has 10 mL of a solution of KOH which has pH = 10. The solution is diluted with de-ionised water to make up a litre of solution. The pH of the diluted solution will be
A. 8
B. 9
C. 11
D. 12

The answer is A but why? This is what I did:

$pH = 10 \rightarrow [H3O^+] = 10^{-10} M$ $\rightarrow$ $C_1V_1=C_2V_2$

$10^{-10}*0.01=C_2$ $\rightarrow$ $C_2=10^{-12}$

$pH =12$

But then i did it another way:

$pH = 10 \rightarrow [H3O^+] = 10^{-10} M$

$[OH^-]=10^-4$ $\rightarrow$ $C_1V_1=C_2V_2$

$10^{-4}*0.01=C_2$ $\rightarrow$ $C_2=10^{-6}$

$[OH^-]=10^{-6}$ $\rightarrow$ $[H3O^+]=10^{-8}$

$pH =8$

Is it because you can't assume $[OH^-] =[H3O^+]$ ????
Title: Re: pH question...
Post by: roly182 on November 12, 2008, 04:35:32 pm: Thats exactly right.. you cant assume that unless the solution is neutral/pure
Title: Re: pH question...
Post by: Mao on November 12, 2008, 05:59:14 pm: the key is, at 25 degrees, $[H_3O^+]\cdot [OH^-] = 10^{-14}$

pH of 12 would be counter-intuitive - diluting a base makes it stronger???
Title: Re: pH question...
Post by: shinny on November 12, 2008, 06:04:55 pm: For questions like this, just stick to using dilution factors. From 10mL to 1L (1000mL), you're diluting by a factor of 100 i.e. $10^{2}$ . Stick this number into the negative log formula for pH, and you'll realise that this represents a pH change of 2. Since you're diluting a base, obviously it's going to go down towards 7, so it'll go from 10 to 8.
Title: Re: pH question...
Post by: lanvins on November 12, 2008, 07:37:28 pm: what would you do if it were made more concentrated?
Title: Re: pH question...
Post by: onlyfknhuman on November 12, 2008, 07:39:10 pm: Quote from: shinjitsuzx on November 12, 2008, 06:04:55 pm
For questions like this, just stick to using dilution factors. From 10mL to 1L (1000mL), you're diluting by a factor of 100 i.e. $10^{2}$ . Stick this number into the negative log formula for pH, and you'll realise that this represents a pH change of 2. Since you're diluting a base, obviously it's going to go down towards 7, so it'll go from 10 to 8.

nice way of thinking (Y)
Title: Re: pH question...
Post by: lanvins on November 12, 2008, 07:46:39 pm: v