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I cant figure out this acid base problem

  1. Mar 4, 2007 #1
    1. The problem statement, all variables and given/known data
    You dissolve an unknown quanity of weak acid HA into 50 mL of water. It only takes 1 mL of 5 M NaOH to completely neutralize the acid. The Ka of the acid is 3.2 x10^-4. What is the pH at the endpoint?

    2. Relevant equations
    Ka * Kb = Kw
    pH + pOH = 14
    p(of whatever) = -log(of whatever)

    3. The attempt at a solution
    Ok, first, we know only 1 mL of 5 M NaOH was used. So, that means there were only 5mol NaOH/L * 1mol OH/1mol NaOH * 1 mL * 1L/1000mL which gives us 0.005 mol of OH were used. If 0.005 mol of OH was used, that means 0.005 mol of HA was neutralized.
    Now, this is where i got messed up.
    The principle reaction of this system is HA + OH = A + H20
    If 0.005 mol of HA was neutralized by the HA, that means 0.005 mol of the A ion remains. We also know that A will ionize in the H20, so let's rewrite the formula. A + H20 = HA + OH
    where there are initially 0.005 mols of A. We know for a fact all of those mols will not ionize to form 0.005 mol of HA, because since it's the conjugate base of a weak acid, it will not go all the way to the other side. So, we use Kb to determine the extent of the reaction. We are given Ka, and we know the relationship Ka * Kb = 1 x 10^-14.
    Solving for Kb, we get 3.125 x 10^-11. In order to use the Kb relation, we need molarity instead of number of mols. Well, we know that there are 0.005 mol of A dissolved in 50 mL of water, so that gives us an initial molarity (before the reaction starts) of 0.1 M. We know that at equilibrium, there will be X M of OH and HA, so molarity of A at equilibrium is the initial molarity - molarity of OH, which is 0.1 - X. We can set the equation like any other reaction

    Kb = ([OH][HA])/A
    3.125 x 10^-11 = (x^2)/(0.1-x)
    Solving for x which equals [OH], we get 2 x 10^-6.
    Since we know [OH], we can find pOH which is -log[OH] = 5.69897
    and we have the handy dandy formula pH + pOH = 14
    pH = 14 - pOH = 8.30103

    Now, i think i did the problem correctly, but i just have this feeling in my gut i did something wrong. Can someone run through my steps, or do the problem yourself and see if u get that answer?
    This is a take home assignment, and i gotta get good grades on the take home assignments. My entier grade in the class is based off of labs, take home assignments, and tests. Now, i can afford to screw up on tests and labs, because we can have retakes on tests, and my teacher lets us do lab corrections. But, there are no second chances on take home assignments.
    thanks alot.
    Last edited: Mar 4, 2007
  2. jcsd
  3. Mar 5, 2007 #2


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    Homework Helper
    Education Advisor
    Gold Member

    This is like asking what is the pH of the conjugate base of that weak acid. How many moles of HA was equivalent? What is the formality of the resulting NaA salt? You already have the Ka value, so take it from there.
  4. Mar 5, 2007 #3


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    Staff: Mentor

    One thing is definitely wrong. What is your FINAL volume?

    At the endpoint you have solution of salt of weak acid. Check pH of weak acid salt lecture.
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