sorry to take up your time again, acids and bases just really confuse me, i'd really appreciate your help again.
"What is the pH at the end point for the titration of 0.20M HNO2 by 0.20M NaOH? Ka nitrous acid= 4.5 x 10^-4."
this is what i do know:
- end point, so it's the salt solution of a weak acid/strong base
- i THINK the formula for the ICE chart is NO2^- + H2O ==> HNO2 + OH^-
- i think i need to use the Ka to solve for x now. is this thinking wrong?
what i'm confused about:
- how do i find the concentration of HNO2, is it 0.20M?
- is the OH- added 0.20M also?
correction for typo: "- how do i find the concentration of HNO2, is it 0.20M? "
i meant NO2^- instead of HNO2
You're on the right track.
............NO2^- + HOH ==> HNO2 + OH^-
initial.....0.1.............0........0
change......-x...............x.......x
equil......0.1-x.............x........x
Kb for NO2^- = (Kw/Ka for HNO2) =
(HNO2)(OH^-)/(NO2^-)
(HNO2) = (OH^-) = x
(NO2^-) = 0.1-x.
Solve for x which is OH, convert that to pH.
Most students ask, "Where did the 0.1M come from for NO2^-?" If we have 0.2M acid titrated with 0.2M base, then the salt formed is 0.2M EXCEPT that it has been diluted. If you do it in moles it is easier to see.
We start with an easy number like 100 mL x 0.2M acid = 20 millimoles acid. That reacts exactly with 20 mmoles base to form 20 mmoles salt. What is the concn of the salt? It is M = mmoles/mL = 20 mmoles/200 mL = 0.1M. You are looking for shortcuts. When you titrate acids with bases and the concns are the same, as they are in this problem, then the concn of the salt is just 1/2 of the starting molarity of the acid (or base). So 0.1M HCl titrated with 0.1M NaOH produces 0.05M NaCl.
0.3M HNO2 + 0.3M NaOH produces 0.3/2 = 0.15M NaNO2.
you're the best, thank you so much!
No problem, I'm here to help! Let's break down the problem and go step by step.
To determine the pH at the end point of the titration, we need to consider the reaction that occurs between the weak acid (HNO2) and the strong base (NaOH). You are correct that the formula for the reaction is:
HNO2 + NaOH -> NaNO2 + H2O
Now, let's address your questions.
1. Concentration of HNO2:
In this problem, it is given that the concentration of HNO2 is 0.20M. So, you can assume the initial concentration of HNO2 to be 0.20M.
2. Concentration of NaOH (OH-):
Since it is mentioned that we are using 0.20M NaOH, it means that the concentration of OH- ions added is also 0.20M.
Now that we have the initial concentrations, we can proceed with the ICE chart and use the Ka of nitrous acid to solve for x.
The balanced equation is:
HNO2 + OH- -> NO2- + H2O
Setting up the ICE table:
Initial:
HNO2: 0.20M
OH-: 0.20M
NO2-: 0M
H2O: 0M
Change:
HNO2: -x
OH-: -x
NO2-: +x
H2O: +x
Equilibrium:
HNO2: 0.20M - x
OH-: 0.20M - x
NO2-: x
H2O: x
The equilibrium expression for this reaction is:
Ka = [NO2-][H2O] / [HNO2][OH-]
However, since water is a pure liquid and its concentration remains constant, we can omit it from the expression:
Ka = [NO2-] / [HNO2][OH-]
We can substitute the equilibrium concentrations into the expression and solve for x.
After solving for x, you can calculate the concentration of H+ ions using the equation:
[H+] = 10^(-pH)
Note that at the end point, HNO2 is completely consumed, and only the conjugate base NO2- is present.
Hope this clarifies things for you!