How to Prepare a Buffer Solution with Methanoic Acid and Methanoate Ions?

[hobomoe]

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.

 

[chemisttree]

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

This is where you went wrong.

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.

What is the target fluoride level? (it doesn't need to be F^-, BTW)

 

[hobomoe]

I got the top one right I'm pretty sure. The bottom one is regarding precipitates, so Qs can't be greater than Ks. I got that the scientist was right but I'm too lazy to type it out.

 

[mcbud]

For the second question, fluorine and calcium can form a precipitate depending on the concentrations of the two ions. Since [Ca²⁺] = 1.86\times10^-3 M, using the K_sp of CaF₂ you can find the equilibrium concentration of the F- anion, which is the maximum concentration that the solution could hold before it precipitates: K_sp = [Ca²⁺][F-]², plug in the the values of K_sp and Ca²⁺, and you should get the [F-]. A fluoride concentration higher than this number will precipitate calcium fluoride.

Another way of solving this problem is using the ion product Q, which has the form of the K_sp, except that the concentrations are the actual or proposed (not equilibrium) concentrations. If Q > K_sp, then the salt will precipitate.