Turn on thread page Beta
    • Thread Starter
    Offline

    1
    ReputationRep:
    0.3M of NaOH.

    Calculate the pH when Kw = 1x10^14 at 298K.


    I dont get how to do this.

    I know that Kw = [H+] x [OH-]

    But since I only know the conc. of the NaOH, I dont understand how to use that identity above to work out the pH of NaOH.

    Or am I being stupid and confusing myself over information that has no bearing on the answer?


    Thanks
    Offline

    19
    use pOH = -log [OH-] then pH + pOH = 14
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by EierVonSatan)
    use pOH = -log [OH-] then pH + pOH = 14
    eh?

    surely its pH = -log(OH), so pH = 0.522

    just like it is for acids?
    Offline

    19
    (Original post by 2strong)
    eh?

    surely its pH = -log(OH), so pH = 0.522

    just like it is for acids?
    nope:

    potential of hydrogen pH = -log [H+]

    potential of hydroxide pOH = -log [OH-]
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by EierVonSatan)
    nope:

    potential of hydrogen pH = -log [H+]

    potential of hydroxide pOH = -log [OH-]
    what!?

    wait hang on.

    If there is 1.4 M of H2SO4, what is its pH? -0.442 right?
    Offline

    19
    (Original post by 2strong)
    what!?

    wait hang on.

    If there is 1.4 M of H2SO4, what is its pH? -0.442 right?
    yeah (well -0.447)
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by EierVonSatan)
    yeah (well -0.447)
    Ok, so how is the method I used to get that answer different to the method I use to get the answer to the question in the opening post?
    Offline

    19
    (Original post by 2strong)
    Ok, so how is the method I used to get that answer different to the method I use to get the answer to the question in the opening post?
    You want to find the pH, and you're given the concentration of NaOH (a strong base so can assume [NaOH] = [OH-]). You cannot find the pH directly from this and so you are given that Kw = [OH-][H+] = 10-14.

    So, firstly find the pOH then use pH + pOH = 14 to find the pH.

    If pH = -log [OH] then > pH = 0.55 is a bit low for a strong base :yep:
    Offline

    13
    ReputationRep:
    (Original post by 2strong)
    0.3M of NaOH.

    Calculate the pH when Kw = 1x10^14 at 298K.


    I dont get how to do this.

    I know that Kw = [H+] x [OH-]

    But since I only know the conc. of the NaOH, I dont understand how to use that identity above to work out the pH of NaOH.

    Or am I being stupid and confusing myself over information that has no bearing on the answer?


    Thanks
    pOH = 13.48
    pH = 0.52

    Add them together to get a pH of 14.

    No?

    PS: It's 1 x 10^-14
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by EierVonSatan)
    You want to find the pH, and you're given the concentration of NaOH (a strong base so can assume [NaOH] = [OH-]). You cannot find the pH directly from this and so you are given that Kw = [OH-][H+] = 10-14.

    So, firstly find the pOH then use pH + pOH = 14 to find the pH.

    If pH = -log [OH] then > pH = 0.55 is a bit low for a strong base :yep:
    The underlines pH and the boldend pH are different arnt they?
    Offline

    19
    (Original post by 2strong)
    The underlines pH and the boldend pH are different arnt they?
    The bottom line is your suggestion, and is wrong

    This is how you do it;

    pOH = -log (0.3) = 0.523

    pH = 14 - 0.523 = 13.5

    so pH = 13.5
 
 
 
Poll
Could you cope without Wifi?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.