Post by: schnappy on November 15, 2010, 08:46:38 pm
I've completely forgotten how to do these easy questions. So... pH = -log([H+])
What's the pH of a 0.001 M sodium hydroxide soln?
Thanks
Post by: stonecold on November 15, 2010, 08:48:44 pm
Therefore [OH-]=0.001M
[H+] = 10^-14/0.001
Hence pH=-log(10^-14/0.001)
Post by: andy456 on November 15, 2010, 08:50:28 pm
-log[OH]
= x
pH=14 - x
Post by: schnappy on November 15, 2010, 09:02:02 pm
Post by: luken93 on November 15, 2010, 09:17:26 pm
Also, I got a couple of questions.
If it's a diprotic such as H2SO4 do you double the [H+] concentration? Can't remember if I'm thinking of the wrong thing...
Also, I had a question today that I can't remember how to do.
250ml of .0500M H2SO4
a) find pH
b) If 1L of water is added,what is the new pH
c) Find the Volume of NaOH needed to equalise the solution?
Post by: stonecold on November 15, 2010, 09:28:42 pm
(Soz to hijack your thread)
Also, I got a couple of questions.
If it's a diprotic such as H2SO4 do you double the [H+] concentration? Can't remember if I'm thinking of the wrong thing...
Also, I had a question today that I can't remember how to do.
250ml of .0500M H2SO4
a) find pH
b) If 1L of water is added,what is the new pH
c) Find the Volume of NaOH needed to equalise the solution?
Yeah, you double the concentration.
a) pH=-log(0.1)=1
b) Find mol H+
n(H+)=cv=0.1x0.25=0.025 mol
Find new conc of H+
c=n/v=0.0025/1.25=0.02
pH=-log(0.02)=1.70
c) H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
Just realised you can't do it without the concentration of the NaOH...
Post by: pi on November 15, 2010, 09:46:36 pm
would this help?
Post by: stonecold on November 15, 2010, 09:47:39 pm
Does pOH = 14 + log10[H+] ???
would this help?
nah, it is just pOH=-log([OH-])
but your way might work, although i've never used it.
Moderator action: removed real name, sorry for the inconvenience
Post by: luken93 on November 15, 2010, 09:51:42 pm
Oh it was 0.1M(Soz to hijack your thread)
Also, I got a couple of questions.
If it's a diprotic such as H2SO4 do you double the [H+] concentration? Can't remember if I'm thinking of the wrong thing...
Also, I had a question today that I can't remember how to do.
250ml of .0500M H2SO4
a) find pH
b) If 1L of water is added,what is the new pH
c) Find the Volume of NaOH needed to equalise the solution?
Yeah, you double the concentration.
a) pH=-log(0.1)=1
b) Find mol H+
n(H+)=cv=0.1x0.25=0.025 mol
Find new conc of H+
c=n/v=0.0025/1.25=0.02
pH=-log(0.02)=1.70
c) H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
Just realised you can't do it without the concentration of the NaOH...
Post by: stonecold on November 15, 2010, 09:54:44 pm
Oh it was 0.1M(Soz to hijack your thread)
Also, I got a couple of questions.
If it's a diprotic such as H2SO4 do you double the [H+] concentration? Can't remember if I'm thinking of the wrong thing...
Also, I had a question today that I can't remember how to do.
250ml of .0500M H2SO4
a) find pH
b) If 1L of water is added,what is the new pH
c) Find the Volume of NaOH needed to equalise the solution?
Yeah, you double the concentration.
a) pH=-log(0.1)=1
b) Find mol H+
n(H+)=cv=0.1x0.25=0.025 mol
Find new conc of H+
c=n/v=0.0025/1.25=0.02
pH=-log(0.02)=1.70
c) H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
Just realised you can't do it without the concentration of the NaOH...
okay, so n(H2SO4)=cv=0.05x0.25=0.0125 mol
n(NaOH)= 2 x n(H2SO4) = 0.025 mol
v=n/c=0.025/0.1=0.25L
Post by: luken93 on November 15, 2010, 10:00:43 pm
yay all right :)Oh it was 0.1M(Soz to hijack your thread)
Also, I got a couple of questions.
If it's a diprotic such as H2SO4 do you double the [H+] concentration? Can't remember if I'm thinking of the wrong thing...
Also, I had a question today that I can't remember how to do.
250ml of .0500M H2SO4
a) find pH
b) If 1L of water is added,what is the new pH
c) Find the Volume of NaOH needed to equalise the solution?
Yeah, you double the concentration.
a) pH=-log(0.1)=1
b) Find mol H+
n(H+)=cv=0.1x0.25=0.025 mol
Find new conc of H+
c=n/v=0.0025/1.25=0.02
pH=-log(0.02)=1.70
c) H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
Just realised you can't do it without the concentration of the NaOH...
okay, so n(H2SO4)=cv=0.05x0.25=0.0125 mol
n(NaOH)= 2 x n(H2SO4) = 0.025 mol
v=n/c=0.025/0.1=0.25L
thanks buddy
Post by: pi on November 16, 2010, 04:55:49 pm
Does pOH = 14 + log10[H+] ???
would this help?
nah, it is just pOH=-log([OH-])
but your way might work, although i've never used it.
I think it does:
pH + pOH = 14
Therefore, pOH = 14 - pH
= 14 - (-log10[H+])
= 14 + log10[H+]
Its a way not in the textbooks though (from what I have seen) ;).
Moderator action: removed real name, sorry for the inconvenience
Post by: schnappy on November 17, 2010, 01:24:52 am
(Soz to hijack your thread)
Also, I got a couple of questions.
If it's a diprotic such as H2SO4 do you double the [H+] concentration? Can't remember if I'm thinking of the wrong thing...
Also, I had a question today that I can't remember how to do.
250ml of .0500M H2SO4
a) find pH
b) If 1L of water is added,what is the new pH
c) Find the Volume of NaOH needed to equalise the solution?
Yeah, you double the concentration.
a) pH=-log(0.1)=1
b) Find mol H+
n(H+)=cv=0.1x0.25=0.025 mol
Find new conc of H+
c=n/v=0.0025/1.25=0.02
pH=-log(0.02)=1.70
c) H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
Just realised you can't do it without the concentration of the NaOH...
Thread hijack? Don't be silly. There was a largeish question on pH/concentrations with a diprotic acid. Finding this out last night I thinks given me a real boost :) Even though it's not important, year 11 and all, I still like to do reasonably well.
Post by: taiga on November 17, 2010, 01:26:08 am
Does pOH = 14 + log10[H+] ???
would this help?
nah, it is just pOH=-log([OH-])
but your way might work, although i've never used it.
I think it does:
pH + pOH = 14
Therefore, pOH = 14 - pH
= 14 - (-log10[H+])
= 14 + log10[H+]
Its a way not in the textbooks though (from what I have seen) ;).
Only at 25 degrees mind you.
Moderator action: removed real name, sorry for the inconvenience
Post by: pi on November 17, 2010, 05:39:30 pm
Does pOH = 14 + log10[H+] ???
would this help?
nah, it is just pOH=-log([OH-])
but your way might work, although i've never used it.
I think it does:
pH + pOH = 14
Therefore, pOH = 14 - pH
= 14 - (-log10[H+])
= 14 + log10[H+]
Its a way not in the textbooks though (from what I have seen) ;).
Only at 25 degrees mind you.
forgot to mention that... yr 11 pH doesn't change with temperature anyway
Moderator action: removed real name, sorry for the inconvenience
Post by: luken93 on November 17, 2010, 05:46:02 pm
haha no worries, happy to help haha(Soz to hijack your thread)
Also, I got a couple of questions.
If it's a diprotic such as H2SO4 do you double the [H+] concentration? Can't remember if I'm thinking of the wrong thing...
Also, I had a question today that I can't remember how to do.
250ml of .0500M H2SO4
a) find pH
b) If 1L of water is added,what is the new pH
c) Find the Volume of NaOH needed to equalise the solution?
Yeah, you double the concentration.
a) pH=-log(0.1)=1
b) Find mol H+
n(H+)=cv=0.1x0.25=0.025 mol
Find new conc of H+
c=n/v=0.0025/1.25=0.02
pH=-log(0.02)=1.70
c) H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
Just realised you can't do it without the concentration of the NaOH...
Thread hijack? Don't be silly. There was a largeish question on pH/concentrations with a diprotic acid. Finding this out last night I thinks given me a real boost :) Even though it's not important, year 11 and all, I still like to do reasonably well.
Post by: ruchika5 on March 04, 2011, 10:20:24 pm
1) What is the pH of a solution produced with 3.5g of nitric acid dissolves in 150mL water?
2) What is the pH of a solution when 70mL of 0.2M HCl is added to 30mL 0.5M barium hydroxide solution?
3)What mass of HCl must be added to 80mL 0.5M potassium hydroxide to lower the pH to 5?
Thanks!
Post by: schnappy on March 04, 2011, 10:51:18 pm
HNO3
n(H+)=n(HNO3)=3.5/63=0.0555556
[H+]=n/v=0.0555556/0.150=0.3703704
pH=-log(0.3703704)=profit
Post by: thushan on March 04, 2011, 10:51:59 pm
So you want to first know the amount of HNO3 per litre. --> need n(HNO3)
n(HNO3) = m/M = 3.5/(1.0 + 14.0 + 16.0 x 3) = 0.056 mol
[H+] = [HNO3] = 0.056/0.150 = 0.37 M (n = cV)
=> pH = -log[10] (0.37) = 0.431
2)
need [H+] or [OH-]
n(OH-)[after reaction] = 0.030 x 0.5 x 2 - 0.070 x 0.2 = 0.016 mol
[OH-] = 0.016/0.100 = 0.16 M
=> pH = 13.2
3)
you want [H+] = 10^(-5) => n(H+) = 8 x 10^(-7)
The odd thing about this question is that the above number, representing the excess [H+], is very very very small. This is basically equivalent to adding equilvalent amount of HCl for neutralisation.
in that case you'd get 1.46 g of HCl. If you really want to be precise to get a pH of 5, you need 1.4600292 g.
Post by: schnappy on March 04, 2011, 10:52:42 pm
Post by: nacho on March 04, 2011, 10:53:58 pm
Post by: thushan on March 04, 2011, 10:56:16 pm
Post by: Mao on March 04, 2011, 11:11:01 pm
you're both lazy bums for not using latex :P
Post by: pi on March 05, 2011, 10:03:29 am
Alternatively
-log[OH]
= x
pH=14 - x
Simplify to pH = 14 + log[OH].
Saves a step