## Weak Acid & Salt

Melody P 2B
Posts: 35
Joined: Fri Sep 29, 2017 7:06 am

### Weak Acid & Salt

I just want to make sure I understand something, for the example we did in class today:

Calculate the pH of solution with 0.100 M HNO2 and 0.15 M KNO2, Ka= 4.3x10^-4

The chemical equation is: HNO2(aq) + H2O (l) (eq. arrows) NO2- + H3O+ because in the chemical equation

(NO2+) + (K-) + (HNO2) (eq. arrows) (NO2-) + (H3O+) + (K-) + (NO2+)

the (K-) and (NO2+) cancel on both sides?

Hovik Mike Mkryan 2I
Posts: 95
Joined: Fri Sep 28, 2018 12:25 am

### Re: Weak Acid & Salt

Hello,
I believe that the potassium onion is present on both sides alone, therefore they have no affect on the concentration of the pH or POH and they cancel as mentioned. But the HNO2 is present only on the left and the deprotonated NO2 is on the right. There isn't another separate NO2 molecule on the left. Hope this helped!

204929947
Posts: 76
Joined: Fri Apr 06, 2018 11:03 am

### Re: Weak Acid & Salt

The Potassium, K, cancels on both sides which is why we don't write it on the equation. We only write NO2-

Niveda_B_3I
Posts: 60
Joined: Fri Sep 28, 2018 12:24 am

### Re: Weak Acid & Salt

since it's a asalt, the potassium ions dissociate completely on both sides, and don't really count toward finding your equilibrium constant value.

Kriti Goyal 4K 19
Posts: 32
Joined: Fri Sep 28, 2018 12:21 am

### Re: Weak Acid & Salt

Because it is a salt, KNO2 will completely dissociate in a solution. Now when you write the reaction both K+ and NO2- show up both sides of the equation, NO2- on the product side does not have the same concentration as the NO2- on the reactant side as it adds up to the concentration of NO2- dissociated from HNO2. but because there is no extra K+ added, its concentration is equal on both sides of the equation.
Now the reason you cancel out the K+ is because when you put it in the expression for Ka, it would cancel out anyway because of the numerator and denominator values being the same for it. This would not hold true for NO2- however.