How do I writing Bronsted equations and determine the equilibrium constants for acid-base reactions?

Question

I seem to keep messing up trying to write the reaction in terms of Acid(1) + Base(2) = Base(1) + Acid(2). On the first reaction I know HF is the acid and therefore F^1- is the conjugate base but I can not put it all together. Below are the reactants I am having problems with. And I have no clue to find K for any of them. I am given this website to help me. hxxp://www.pearsoncustom.com/wertz/a_cd/caqs/html/resource/acidbase_pka.html.


Reactants:
1. hydrofluoric acid + potassium cyanide
2. hydrochloric acid + sodium hydroxide
3. formic acid + sodium nitrite
4. KHCO3 + KCN
5. H2SO3 + K2SO4

Please help me and explain. Thank You.

Answer 1

I'll do the first one for you and you can do the rest.

HF + KCN <=>KF +HCN

K+ is a spectator ion, so let's leave it out. Thus we have

HF + CN- <=> F- +HCN

so K= [F-][HCN] / ([HF)[CN-])= ([F-]/[HF]) * ([HCN]/[CN-]) (1)

but from the dissociation of each acid we have

Ka(HF)= [H+][F-]/[HF] => [F-]/[HF]= Ka(HF)/[H+] (2)

Ka(HCN)= [H+][CN-]/[HCN] =>
[HCN]/[CN-] = [H+]/Ka(HCN) (3)

Substitute the ratios in (1) using (2) and (3)

K= (Ka(HF)/[H+]) * ([H+]/Ka(HCN))= Ka(HF)/Ka(HCN)

now plug-in the values for the Ka that they give you and you can find K

Answer 2

There are a number of issues in this question. First, settle on even if the substance is an acid or a base. even if that's ionic, you may want to split the ions earlier you're making that decision (so in f, case in point, evaluate sodium ion and carbonate ion--is both acid or base? if so, basically use the ion it truly is acid or base and ignore about the different) even if that's an acid, then the eqn is obtained by ability of creating use of water as a base. considering that an acid provides up a hydrogen ion and a base features one, the substance it truly is an acid loses H+ and the water features one. If the substance is a base then a H+ is transferred from water to the bottom. get at the same time: assume you necessary to write down this manner information superhighway eqn for the acid HF. considering that HF is an acid, that's going to lose the H+ and form F-. The water then features the H+ and makes H3O+. HF + H2O --> F- + H3O+ Be particularly careful about the expenditures. The acid loses a + to form the conjugate base and the bottom includes a + fee to form the conjugate acid.

Answer 3

In the Bronsted-Lowry theory of acids and bases, acids are proton donors, and bases are proton acceptors. There are no acid(1) or acid(2). Rather, the Bronsted acid transfers a proton to a Bronsted base to form a conjugate acid (the product resulting from having received the proton) and the conjugate base (the product from having lost the proton.)

1. HF (Bronsted acid) + KCN (Bronsted base) ===> HCN (conjugate acid) + KF (conjugate base)

2. HCl (Bronsted acid) + NaOH (Bronsted base) ===> H2) (conjugate acid) + NaCl (conjugate base)

3. HCO2H + NaNO2 ===> HNO2 (nitrous acid, conjugate acid) + HCO2Na (conjugate base).

4. KHCO3 (Bronsted acid) + KCN (Bronsted base) ===> HCN (Conjugate acid) + K2CO3 (Conjugate base.

5. H2SO3 (Bronsted acid) + K2SO4 (Bronsted base) ===> KHSO4 (Conjugate acid) + KHSO3 (conjugate base.