Dr. Bob222:
There is a problem you answered .100M HClO2 titrated with .080M NaOH
I can work the problem many times but I am getting 7.30pH and the answer says 7.38 using quad. I used quad and still get 7.30.
I also do know how you go from the neutralization reaction of HClO2 + NaOH-->???
And then the answer says ClO2 turns into ClO and hydrolized from there?
Can you please assist, thsi problem is driving me nuts.
Thank you and thanks for your assistance.
S
If possible could you also please show the neutralization reaction and the next reaction that follows where ClO2 is involved.
I get X = 2.00896X 10-^7? is that correct
pOH from there 6.70
pH at 7.30 but again my text is saying 7.38 with quadratic.
thanks!
I'm afraid I can't be much help.
1. I worked this from scratch tonight and I also come out with pOH = 6.70 and pH = 7.30 (with or without the quadratic).
2. I searched the old files and found a number of posts I had made for this same problem but none of them gave an answer; i.e., I just worked through the hydrolysis etc.
3. I found one old post that asked for NaOH and chlorous acid(HClO2); however, I worked the problem (again worked meaning just the steps) using HClO and not HClO2. Did that reference to form HClO and then hydrolyzed from there come from my post or from your test?
4. Where did this problem originate?
Chemistry 12th Ed. problem 17.48b
I just wasn't sure either how HClO2 got to ClO2 and which Ka you use?
There was no ref to form HClO
minus the issue on 7.38 if you could comment furhter on the neutralization reaction or the following one, Id like to understand that better. Thanks for getting back to me.
You seem to be the chem guru
Thanks but I'm no guru; Bob Pursley is the all around guru.
HClO2 + NaOH ==> NaClO2 + H2O for the neutralization.
At the equivalence point the pH is determined by the hydrolysis of the
NaClO2 or Kb for ClO2^-.
ClO2^- + HOH ==> HClO2 + OH^- is the hydrolysis reaction for which Kb = Kw/Ka and for Ka I used 1.1E-2. The use of different Ka values usually is the reason for numbers not quite agreeing; however, since both of us arrived at the same answer we must have used Ka values that were the same or at least quite similar.
I understand that you are having trouble with a pH calculation problem involving the titration of HClO2 with NaOH. Let's break down the problem and address your concerns step by step:
1. First, let's discuss the neutralization reaction:
HClO2 + NaOH → NaClO2 + H2O
In this reaction, HClO2 reacts with NaOH to form NaClO2 (sodium chlorite) and water (H2O).
2. Now, let's focus on the pH calculation. To determine the pH of the resulting solution, we need to consider the concentration of the acidic and basic components and how they react.
3. You mentioned using quad, which suggests that you are using the quadratic formula to solve an equation involving an equilibrium position. However, in this titration problem, we do not need to use the quadratic formula because the concentrations of the reactants are not equal.
4. What we can do is calculate the number of moles of the acid and base to determine the limiting reactant. From there, we can determine the remaining concentration of the excess reactant and calculate the final pH.
To do this, follow these steps:
a. Calculate the number of moles of the acid:
Moles of HClO2 = volume (in liters) × concentration of HClO2 (in moles/liter)
b. Calculate the number of moles of the base:
Moles of NaOH = volume (in liters) × concentration of NaOH (in moles/liter)
c. Compare the stoichiometry of the reaction:
The molar ratio of HClO2 to NaOH from the balanced equation is 1:1. Determine if the number of moles of NaOH is equal to or greater than the moles of HClO2. If the moles of NaOH are greater, NaOH is in excess. If the moles of HClO2 are greater, HClO2 is the limiting reactant.
d. Calculate the remaining concentration of the excess reactant:
If NaOH is in excess, subtract the moles of HClO2 reacted from the total moles of NaOH to find the moles of NaOH remaining. Divide this value by the total volume (in liters) to get the new concentration of NaOH (which will be lower).
e. Calculate the concentration of NaClO2:
Divide the moles of NaClO2 formed by the total volume (in liters) to determine the concentration of NaClO2.
f. Calculate the concentration of OH- ions:
Since NaOH is a strong base, it dissociates completely in water to form Na+ and OH- ions. The concentration of OH- ions will be the same as the concentration of NaOH (new concentration from step d or the initial concentration).
g. Calculate the concentration of HClO2:
Subtract the moles of HClO2 reacted (calculated in step c) from the total moles of HClO2 to find the moles of HClO2 remaining. Divide this value by the total volume (in liters) to get the new concentration of HClO2 (which will be lower).
h. Use the concentration of HClO2 to calculate the concentration of H3O+ ions:
Since HClO2 is a weak acid, it does not fully dissociate, and we need to consider the equilibrium expression. You would need the Ka value of HClO2 (acid dissociation constant) and solve the equilibrium expression to find the concentration of H3O+ ions. This is the step where you would need to use the quadratic formula if required.
i. Calculate the pH:
pH=-log[H3O+]. Take the negative logarithm (base 10) of the concentration of H3O+ ions to find the pH of the solution.
I hope this process helps you to rework the problem. If you have any further questions or need more assistance, please let me know!