Question

Thermodynamics of the dissolution of borax report sheet. Need help please !

Pantner's Name Angela Partner's Name Thermodynamics of the Dissolution of Borax Report Sheet Part A Sample number Volume of sample (mL.) 5.00 0 5.00 5.00 2-0 Temperature of sample (°C) Temperature of sample (C)1.la .0 5.0 2 Part B [HCI] (M) Sample number Initial buret reading (mL) Final buret reading (mL) Volume of HCl added (mL)|r (m) .5 4.3 21.2 -LS|HE12 ㄐ 丨1.3 7. 35 m Average volume of HCl at a 1.2 °C(mL) Average volume of HCI at lC (Report this temperature and volume to your instructor) Average volume of HCI at (Report this temperature and volume to your instructor) Part C oC (mL) Temperature (K) Volume of HCI added (mL) (from class data) T (K-) Moles of HCI used Moles of BOs(OH) 81

Answer 1

I will tell you how you'll do to solve this and some examples, so you can fill the table yourself. I'll do examples with 1 trial.

To get the temperature in Kelving, just sum 273 or 273.15 to the temperature in Celsius:
T1 = 21.2 + 273.15 = 294.35 K

To get the volume of HCl, you already have that in part B, just use that volume again there.

1/T, just divide 1 by the temperature you got in Kelvin:
1/294.35 = 3.397x10^-3 K^-1

you have the volume and concentration of HCl, so the moles used:
moles HCl = 0.2 * 0.0074 = 1.48x10^-3 moles

The general reaction is as follow:
B₄O₅(OH)₄ + 2 HCl + 3 H₂O = 4 H₃BO₃ + 2 Cl

According to this reaction, 1 mole of boraz reacts with 2 of HCl, so the moles of borax are:
moles borax = moles HCl / 2 = 1.48x10^-3 / 2 = 7.4x10^-4 moles

To get the concentration of borax just divide the moles between the volume:
C = 7.4x10^-4 / 0.005 = 0.15 M
This is the same molar solubility

Now to get the Ksp, we know that borax:
Na₂B₄O₅(OH)₄ <--------> 2Na⁺ + B₄O₅(OH)₄^2-

If the above concentration is from the borax ion, the sodium is the double so:
[Na] = 0.15*2 = 0.30 M

Ksp = [Na]²[Borax] = 0.30² * 0.15 = 0.0135

and the lnKsp = -4.31

now, we know that $\small \Delta G$ = -RTlnK and $\small \Delta G$ = $\small \Delta H$ - T $\small \Delta S$ ; replacing both in one equation:
$\small \Delta H$ - T $\small \Delta S$ = -RTlnK

Solving for lnK:
-lnK = $\small \Delta H$ /RT - T $\small \Delta S$ /RT
lnK = $\small \Delta S$ /R - $\small \Delta H$ /R(1/T)

So you need to plot lnKsp vs 1/T after you got the values from trial 1, 2, 3, and 4.
The slope will give you dH/R, and the y intercept dS/R.

From there solve for dH and dS, and finally calculate dG with these two values.

Hope this helps