Calculating pH

[schnappy]

Hey,
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

[stonecold]

[NaOH]=0.001M


Therefore [OH-]=0.001M

[H+] = 10^-14/0.001

Hence pH=-log(10^-14/0.001)

[andy456]

Alternatively
-log[OH]
= x

pH=14 - x

[schnappy]

Thanks! I was multiplying and not dividing... hence wrong index law. I see what I did now, thanks

[luken93]

(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?

[stonecold]

(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...

[pi]

Does pOH = 14 + log₁₀[H+]   

would this help?

[stonecold]

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

[luken93]

(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...

Oh it was 0.1M

[stonecold]

(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...

Oh it was 0.1M

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

[luken93]

(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...

Oh it was 0.1M

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

yay all right
thanks buddy

[pi]

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

[schnappy]

(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.

[taiga]

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

[pi]

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