Example of pKa and pH problem.

[Dhamnekar Winod]

Example pKa and pH Problem
Find [H⁺] for a solution of 0.225 M NaNO₂ and 1.0 M HNO₂. The Ka value (from a table) of HNO₂ is 5.6 e-4. (In table, It is 7.4e-4)

pKa = −log Ka = −log(7.4e−4) = 3.14

pH = pka + log ([A⁻]/[HA])

pH = pKa + log([NO₂-]/[HNO₂])

pH = 3.14 + log(1/0.225)

pH = 3.14 + 0.648 = 3.788

\([H^+] = 10^{-pH}= 10^{−3.788} = 1.6×10^{−4}\)

Are the above computation correct?https://www.thoughtco.com/the-ph-and-pk ... ng%20acid.

[ChenBeier]

Should be ok

[Dhamnekar Winod]

In the question, the concentration of NaNO₂ is 0.225 M and that of HNO₂ is 1.0 M.

But in the handerson-Hasselbalch equation, concentration of HNO₂ is 0.225 and of NO₂⁻ , it is 1.0 M. How is that?

I think the reaction is NaNO₂ + HNO₂ → NaOH + NO + NO₂

[ChenBeier]

No HNO₂ is the acid and NaNO₂ is a salt. There is no reaction together. You can use the acid and add some NaOH to neutralize to get the salt.

[Dhamnekar Winod]

I don't understand how does Log(NO₂⁻ / HNO₂) = Log(1/0.225 )?

Henderson- Hasselbalch equation states pH= pKa + log([conjugate Base ]/[ weak acid])

In this case, NO₂⁻ ia a conjugate base and its concentration is unknown. How to find out that? The concentration of HNO₂ is 1.0 M. But in ths solution it is 0.225 M. How is that?

[ChenBeier]

Something has to be given pH either a concentration. Otherwise it can not work. You have a acid given 1 M and some salt given 0.225 M then you can calculate the pH.
Or pH is given and concentration of the salt or acid can be calculated.
But I see you mixed concentrations up in the beginning.

[Dhamnekar Winod]

Whatever is given on https://www.thoughtco.com/the-ph-and-pk ... ng%20acid. , I reproduced here.

Consider the reaction HNO₂ + NaOH → NaNO₂ + H₂O , Now we have 1.0 M of HNO₂ and 0.225 M of NO₂⁻.

So, pH= 3.13 + log (0.225/1) = 3.13 -0.648 = 2.48 . → [H⁺] = 10^-2.48 =3.31e-3.

Why is this computation wrong?

[ChenBeier]

This is correct but in the beginning you mixed up the concentrations.