# A2 hard acid-base question

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#1
a solution is made by adding 25 cm of 0.1M sodium hydroxide solution t0 24.5 cm of 0.1 hydrochloric acid. what is the final PH of the resulting solution
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#2
im up to this bit
0.5 cm of Na0H in 0.1 mol dm (excess)
so 1.01 x 10^-3 mol dm cubed so whats next?
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3 months ago
#3
(Original post by interlanken-fall)
im up to this bit
0.5 cm of Na0H in 0.1 mol dm (excess)
so 1.01 x 10^-3 mol dm cubed so whats next?
Next, you find the concentration of H+ and then pH can be found
1
3 months ago
#4
(Original post by interlanken-fall)
a solution is made by adding 25 cm of 0.1M sodium hydroxide solution t0 24.5 cm of 0.1 hydrochloric acid. what is the final PH of the resulting solution
You have found the excess of NaOH, but you need the number of mol. (M x vol)
You know the total volume so you can determine the concentration (mol/dm3)
With the concentration of OH- ions, you can use the kw of water (1 x 10-14) to work out the [H+] and from that, the pH
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#5
(Original post by charco)
You have found the excess of NaOH, but you need the number of mol. (M x vol)
You know the total volume so you can determine the concentration (mol/dm3)
With the concentration of OH- ions, you can use the kw of water (1 x 10-14) to work out the [H+] and from that, the pH
(Original post by deskochan)
Next, you find the concentration of H+ and then pH can be found
I used a different way
I found the limiting and excess and the change was = 0.00005 (excess-limiting)
and did that over (24.5+25/1000) and found it to be 1.01 x 10-3
did
-log10(1.01 x 10-3) got 2.99, is this right for the pH?
my teacher did 10^(14-2.99) and got 1.02 x 10^11, why did she do the last bit? i thought we were finding the PH
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#6
(Original post by charco)
You have found the excess of NaOH, but you need the number of mol. (M x vol)
You know the total volume so you can determine the concentration (mol/dm3)
With the concentration of OH- ions, you can use the kw of water (1 x 10-14) to work out the [H+] and from that, the pH
whats kw of water (1 x 10-14)?
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3 months ago
#7
(Original post by interlanken-fall)
I used a different way
I found the limiting and excess and the change was = 0.00005 (excess-limiting)
and did that over (24.5+25/1000) and found it to be 1.01 x 10-3
did
-log10(1.01 x 10-3) got 2.99, is this right for the pH?
my teacher did 10^(14-2.99) and got 1.02 x 10^11, why did she do the last bit? i thought we were finding the PH
-log10(1.01 x 10-3) got 2.99, is this right for the pH?
It is pOH = -log10 [OH-], pH +pOH = 14, thus pH = 14 - pOH = 14-2.99 =11.01
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#8
(Original post by deskochan)
-log10(1.01 x 10-3) got 2.99, is this right for the pH?
It is pOH = -log10 [OH-], pH +pOH = 14, thus pH = 14 - pOH = 14-2.99 =11.01
im confused on the last bit, we found the pOH as NaOH was the excess
if HCl was excess thenw e wouldn't have to do the 14- X
but as its NaOH we are trying to find the p0H of it right?
if so then what does pH and pOH mean?
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3 months ago
#9
(Original post by interlanken-fall)
im confused on the last bit, we found the pOH as NaOH was the excess
if HCl was excess thenw e wouldn't have to do the 14- X
but as its NaOH we are trying to find the p0H of it right?
if so then what does pH and pOH mean?
log10(1.01 x 10-3) got 2.99. It is excess [OH-] and pH value should be greater than 7 but now it shows 2.99. Thus, 2.99 does not represent pH value but pOH because we use the concentration of OH-, not H+. That is we can use two directions to solve it, 1. use [H+][OH-] = 10^-14 to get [H+] and then pH value. 2. pH+pOH=14. You can read this https://courses.lumenlearning.com/ch...e-poh-concept/
Last edited by deskochan; 3 months ago
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#10
(Original post by deskochan)
log10(1.01 x 10-3) got 2.99. It is excess [OH-] and pH value should be greater than 7 but now it shows 2.99. Thus, 2.99 does not represent pH value but pOH because we use the concentration of OH-, not H+. That is we can use two directions to solve it, 1. use [H+][OH-] = 10^-14 to get [H+] and then pH value. 2. pH+pOH=14. You can read this https://courses.lumenlearning.com/ch...e-poh-concept/
ohhhhhh so if acid was the excess and we got 2 we couldn't need to 14-2
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3 months ago
#11
(Original post by interlanken-fall)
ohhhhhh so if acid was the excess and we got 2 we couldn't need to 14-2
The point is if acid was excess, you can directly get [H+] and pH value and show less than 7.
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#12
(Original post by deskochan)
The point is if acid was excess, you can directly get [H+] and pH value and show less than 7.
0
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