Question

Now, work backwards. If a 0.098 M weak acid solution has a pH of 4.51, calculate the acid ionization constant, Ka

Answer 1

- 0.098 m weak acid solution has a Pr = 4.51 ka = ? -Aid ionization constant. I'm assuming that you are working with mono protic weak acid here so that the equilibrium can be written like this: a ionisation A caq) + Hz ot coq) HA cay) + Hoo (e) ? Now, the solution has [pH = 4.51 The PM of the solution pH = -logo. Chor) - is defined as lothlorog, hol] -pH = eng [hot ] - a Mgot] = loen I Hz ot ) - 10-4:51 Shoot ] = 0.0000 309 [1204) - 3.09 x10-51

Now, notice that every mole of HA that ionizes produces I mole of a the contugate base ay the acid, and I mole of hydronium cations. This means that, at equilibrium the solution has ratio TAJ: [H₂ot J - produced in a 1:1 mole DSCA) = 3109x10-5 The initial concentration af acid will decrease because some of the mole wees ionized to produce Ar and ₂ ot. so, in order og ionization to produce [lgot]. the initial concentration of acid must decrease by [H307]. This means that, at equilibrium, the concentration of the weak deid will be equal to [HA] = 1HAT insihol - [Mgo? ] THAJ = 0.098M - 3.09 tio's m [HA] = 0.0979 M y the dissociation constant ka, will be equal to Kas [A] [Mgot] гня) 1. Ka = 3.0981005 x 3.09 Xions - 34.75X70M 0.0979