Consider the reaction IO−4(aq)+2H2O(l)⇌H4IO−6(aq);Kc=3.5× 10−2IO4−(aq)+2H2O(l)⇌H4IO6−(aq);Kc= 3.5×10−2 If you start with 25.0 mLmL of a 0.909 MM solution of NaIO4NaIO4, and then dilute it with water to 500.0 mLmL, what is the concentration of H4IO−6H4IO6− at equilibrium?

Answer asap need to be answered by wednesday morning explain how a buffer works, using an ethanoic acid / sodium ethanoate system including how the system resists changes in ph upon addition of a small amount of base and upon addition of a small amount of acid respectively. include the following calculations in your i. calculate the ph of a solution made by mixing 25cm3 0.1m ch3cooh and 40cm3 0.1m ch3coo-na+. [ka = 1.74 x 10-5 m] ii. calculate the ph following the addition of a 10cm3 portion of 0.08 m naoh to 500cm3 of this buffer solution. iii. calculate the ph following the addition of a 10cm3 portion of 0.08 m hcl to 200cm3 of the original buffer solution.