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Homework Help: Im so lost in this Problem

  1. Jun 15, 2006 #1
    Can someone direct me with the steps that are needed to solve this question thank you very much for your help!

    What is the pH of a solution resulting from the mixing of 100 mL of 0.250 mol•L-1 KOH solution with 200 mL of 1.20 mol•L-1 HNO3 solution?
  2. jcsd
  3. Jun 15, 2006 #2


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    First write out the balanced equation you predict will occur.

    Since KOH and a strong base and HNO3 is a strong acid, you have an acid base neutralization reaction on your hands.
    There are 3 possibilities of what will happen....
    -The KOH and HNO3 will completely neutralize each other, neither in excess, resulting in a neutral solution,
    -There is a higher number of moles of KOH than necessary to neutralize the HNO3, thus an alkaline solution will result,
    -There is a higher number of moles of HNO3 than necessary to neutralize the KOH, thus an acidic solution will result

    You need to calculate the number of moles of each, KOH and HNO3, in the reaction.
    Since Molarity concentration is in terms of moles per liter and you know the concentration and the volume, you should easily be able to solve for the number of moles.

    Now that you know the # of moles of each reactant, which is the limiting reactant based on the Stoichiometry of the neutralization reaction?
    How many moles of KOH are required to neutralize 1 mole of HNO3?
    Once you know the excess reactant, you know what and how much will be left over in the end.

    pH = -log ([H+])
    pOH = -log (OH-)

    pH + pOH = pKw = 14

    So if you have excess HNO3, the solution will be acidic and you can calculate the pH direction from the number of excess moles of HNo3.
    If the excess reagent is KOH, you can use this to calculate the pOH, and from this, the final pH of the solution.

    Remember, when mixing these two solutions (KOH and HNO3), the volume are added together, take this into account when calculating the final molarity of the excess reagent.
  4. Jun 15, 2006 #3
    Wow Thanks alot i Understand alot now!!
  5. Jun 15, 2006 #4
    Can you also help me out on this other question THANK YOU VERY MUCH!!
    What volume of 0.100 mol•L-1 NaOH (aq) is required to neutralize

    b) 0.250 g of benzoic acid, C6H5COOH? (a weak acid)
  6. Jun 16, 2006 #5


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    Okay, you know the concentration of the sodium hydroxide and thus you know how many hydroxide ions it will put into solution. To neutralise the base, you must put an equal number of protons into solution. Therefore, you will need to calculate the number of protons that benzoic acid will but into solution; you can do this using the acid dissociation constant (Ka) for benzoic acid. To do this you must make two assumtions;

    Assumption One

    [itex][H^{+}_{(aq)}] = [A^{-}_{(aq)}][/itex], (where A- is the ionic salt) this means that all the protons in solution originate from the weak acid. In otherwords, we ignore the ionisation of water (as this is negligable).

    Assumption Two

    When we calculate the concentration of the acid (HA) at equilibrium we assume that no dissociation has occured, this is valid since weak acids dissociate very little.

    Now, can you go from here? Using the two assumptions and the expession for Ka (you will need to look up Ka for benzoic acid). These to me look like homework questions to me, if so could you please post such questions in the homework sections in future. Thank you.
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