Find the ph of a solution that is 0.12 m in nitrous acid and 0.15 m in sodium nitrite. answer in units of ph. 027 (part 2 of 3) 10.0 points now we add 0.15 mol of naoh to 1.00 liter of the solution that was given in part 1. what is the ph of the solution after the naoh addition? answer in units of ph. 028 (part 3 of 3) 10.0 points now, you’re given only 500 ml of the buffer described in part 1 above. to this, 124 ml of 0.50 m hcl is added. what is the new ph of this new solution? answer in units of ph.

Answer the questions about this reaction: nai(aq) + cl2(g) → nacl(aq) + i2(g) write the oxidation and reduction half-reactions: oxidation half-reaction: reduction half-reaction: based on the table of relative strengths of oxidizing and reducing agents (b-18), would these reactants form these products? write the balanced equation: answer options: a. 0/na -> +1/na+1e- b. nai(aq) + cl2(g) → nacl(aq) + i2(g) c. +1/na+1e- -> 0 /na d. -1/2i -> 0/i2+2e- e. no f. 4nai(aq) + cl2(g) → 4nacl(aq) + i2(g) g. 2nai(aq) + cl2(g) → 2nacl(aq) + i2(g) h. 4nai(aq) + 2cl2(g) → 4nacl(aq) + 2i2(g) i. nai(aq) + cl2(g) → nacl(aq) + i2(g) j. 0/cl2+2e -> -1/2cl- k. yes