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#### asdf123

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##### Help with calculations
« on: June 22, 2024, 01:41:53 AM »
Hi, at the university we received a few calculations, but I am not very good at the following ones. I would be very grateful if someone could check my solutions and write down the solution process. Thank you very much.

(I apologize for any possible language inconsistencies in the assignments – they were not originally in English.)

1) 10 ml of RCOOH solution with a pKa = 6 and a concentration of 0.020 M is mixed with 10 ml of organic solvent, the distribution constant is 3.0. If the pH is set to 6, the concentration of acid in the aqueous phase is 0.012 M. What will be the concentration of acid in the aqueous phase if the pH of the aqueous phase is set to 7?

2) Na-ISE was used to measure the concentration of (Na+) ions. First, 10.00 ml of the sample was analyzed and the voltage of the cell was measured at -246 mV. After adding 1.00 ml of a 20 mM solution, the voltage changed to -199.0 mV. Calculate the molar concentration of Na+ in the sample.

3)
F-ISE is used to determine fluorides in water. For a solution of 1.00x10^-3 M standard, the potential was measured at -211.3 mV, for a 4.00x10^-3 M solution the potential was -238.6 mV. For a sample with unknown potential, it was -226.5 mV. Calculate the molar and mass concentration of fluorides in the sample? Ar(F-) = 18.998

4)
A sample of purified organic acid with a mass of 8.90 mg was dissolved in a mixed solvent sample of water-ethanol and was titrated coulometrically with hydroxide ions generated by electrolysis with a current of 44.1 mA for 266 s. Calculate the molecular weight of the acid.

5) Calculate the first equilibrium hydrolytic constant for the reaction Al(H2O)6^3+ ↔ [Al(OH)(H2O)5]^2+ + H+, occurring in an aluminum salt solution with a concentration of 0.001 mol/l, if the equilibrium electrochemical potential of the cell formed by the saturated calomel electrode (E0 = 0.241 V) and the glass electrode is 0.4776 V.

Here is what I tried:

1) I really don't know how, but I am sure, that I would use here somehow the formula: D=(Kd*c(H3O+))/(Ka+c(H3O+)

2) C(mix)=(Cna*Vna + Cst*Vst)/Vtotal
E=E2-E1 = -199-(-246)=47 mV
E=0.0592*ln(Cmix/Cna)
Cmix=6.23Cna

6.23Cna=(Cna*10+30*1)/11
Cna= 6.34 mM

3) y =ax+b
-211.3 = x*10^(-3) + b
-238.6 = x*4*10^(-3) + b

y=-9100x - 202.2
-226.5=-9100x -202.2
x= 6.26*10^(-3)

4) 2H2O + 2e ----- 2OH- + H2

n(OH) = (Qt)/(zF) = (44.1*10^(-3)*266)/(2*96485) = 606*10^(-6) mol
n(acid)= n(OH)
M=m/n = 146.73 g/mol

5) E = E(0) + 0.0592*log (Cox/Cred)
0.4776-0.241=0.0592*log(Cox/Cred)
Cox=9922.5*Cred

C=Cox + Cred
0.001=Cox + Cred

0.001 = 9922.5Cred+ Cred
Cred=1*10^(-7)

K=((Al2+)*(H+))/(Al3+) = (red)^(2)/(0.001 - (red)) = 1*10^(-11)

#### Borek

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##### Re: Help with calculations
« Reply #1 on: June 22, 2024, 04:30:43 PM »
10 ml of RCOOH solution with a pKa = 6 and a concentration of 0.020 M is mixed with 10 ml of organic solvent, the distribution constant is 3.0. If the pH is set to 6, the concentration of acid in the aqueous phase is 0.012 M.

Perhaps I am missing something, but these numbers don't add up. At pH=6 and with pKa=6

[RCOOH] = [RCOO-]

so

[RCOOH] = [RCOO-] = 0.012/2 = 0.006 M.

From the mass balance (initial amount of acid vs amounts of acid in both phases):

10 mL * 0.02 M = 10 mL * 0.012 + 10 mL * x

where x is concentration in the organic phase, so the concentration in the organic phase is 0.02 - 0.012 = 0.008 M.

0.008 M in organic solvent vs 0.006 RCOOH in water (RCOO- is not part of the distribution equilibrium) doesn't yield distribution constant of 3.
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