I had these two questions in a test and I was completely stumped. 1. Make a 1-2L buffer solution containing methanoic acid and methanoate ions and with a pH of 3.74 using 1.07molL-1 HCOOH, solid HCOONa, 1.04molL-1 HCl, 1.16molL-1 NaOH(not all together). pKa(HCOOH)=3.74 M(HCOONa)=68gmol-1 Discuss two methods to make the buffer solution. pH=pKa so HCOOH mol= HCOO- mol 1L HCOOH=1.07mol 1.07 mol of HCOONa x 68gmol-1=72.76g 1L of HCOOH with 72.76g of HCOONa will make a 1L buffer of 3.74 pH. NaOH+HCOOH=HCOONa+H2O 1.07mol NaOH+ 2.14mol HCOOH=1.07mol HCOONa Since the remaining 1.07mol of HCOOH reacts with the 1.07mol of HCOONa no more solutions need to be added. 0.4612L NaOH+ 1L HCOOH=1.4612L buffer solution of 3.74pH. I think this one is correct. 2. F- ions are added to a towns water supply as they believe it helps with dental health. The towns water supply is 'hard' (high conc. of Ca2+ ions) with a conc. of 1.86x10^-3 molL-1. They think a F- conc. of 2.00x10^-4 molL-1 would be appropriate. A scientist recommends 1.31x10^-4 molL-1 because of the calcium ions present. Determine whether the scientist is correct and justify your answer. I have no idea where to begin. I'm guessing it has something to do with the reactions quotient, Qs, and the solubility constant, Ks.