1: Given that Ka for HCOOH is 1.8×10^-4 at 25°C what is the value of b for COOH− at 25 °C? Kb=
2: Given that b for (CH3)2 NH is 5.4×10^-4 at 25°C, what is the value of Ka for (CH3)2NH+2 at 25 °C?
Ka=

Explanation:

At 25 °C Dissociation constant of water or Kw = 1× 10^-14 mol2dm-6

Ka × Kb = Kw for all diluted solutions

Ka×Kb = 1 × 10^-14 mol2dm-6

Therefore for question 1 -

COOH- is the conjugate base for HCOOH ( we find the conjugate base by removing an H atom and the Conjugate acid by adding an H atom)

COOH- (aq) + H2O ( l) <> HCOOH( aq) + OH-(aq)

So we can apply the above mentioned equation to find Kb value for COOH-

Kb = 5.56 × 10^-11 moldm-3

For question 2-

We can see that (CH3)2NH+2 is the conjugate acid of (CH3)2 NH. (A H atom has been removed so the one with a H atom less is the conjugate base.)

(CH3)2NH+2 (aq) + H2O (l) <> (CH3)2 NH (aq) + H3O+ (aq)

So applying the previously mentioned equation we get Ka = 1.85 × 10^-11 moldm-3

2fe + 3h2so4 -> fe2(so4)3 + 3h2

it is necessary to balance equations in order to get the proper amount of reactants as well as products in order not to make a miscalculation.

you cannot change the subscripts as that would change the substance itself. for instance, diatomic elements such as h2 only have the subscript two when a gas. another example would be so4, if you were to change the subscript, the element would no longer be sulfate.

we change the coefficients instead as that only changes the number of molecules rather than the substance itself.

hope this !