5) Using the following data 4) 35.0 ml a Cu 2 volumetric solution of unknown con

Question

5) Using the following data 4) 35.0 ml a Cu 2 volumetric solution of unknown concentration was diluted in a 100.0 ml to flask. The resulting so was analyzed by spectroscopy, and it was found have an trendline to ency of information along with the equation for the solve for the concentration of the original unknown solution. Copper (II) Sulfate standards 2.4 2.2 y 11.401x +0.1264 2.1 19 1.6 1.4 1.2 0.8 0.7 0.6 0.5 0.4 0.3 0.1 o 0,01 0.02 0.03 0,04 0,05 0.06 0.07 0.08 0.09 0.1 0.11 0.12 0.13 0.14 0.150.16 0.17 0.18 0.19 0.2 0.21 Concentration (M)

Explanation / Answer

For the first problem, we have the expression:

y = 11.401x + 0.1264

Where y is the absorbance and x is concentration in M.

We can get first the concentration of new solution with the absorbance given:

0.88 = 11.401x + 0.1264

x = (0.88 - 0.1264) / 11.401 = 0.0661 M = concentration of new solution

Now, we can use dilution equation to get original concentration:

V1M1 = V2M2

35 ml * M1 = 100 ml * 0.0661M

M1 = (100 ml * 0.0661M) / 35 ml = 0.1889 M

Now for the second problem, solving the steps:

a. We can get mass of solution subtracting the mass of calorimeter and contents minus calorimeter mass:

Mass of solution: 103.474g - 3.748g = 99.726 g

b. We obtain change in temperature of the solution:

deltaT = T final - T initial (average) = 28.5 - 21.9 = 6.6oC

c. For moles of limiting reactant, we need to know the reaction:

HCl + NaOH -> NaCl + H2O

So the limiting reactant will be the one with less quantity:

Moles of HCl = 1.5 mol / L * 0.05 L = 0.075 moles

Moles of NaOH = 1.5 mol / L * 0.051 L = 0.0765 moles

So limiting reactant is HCl, with 0.075 moles

d. To get heat of solution:

qsolution = mCpsoldeltaT = 99.726 g * 4.02 J / goC * 6.6oC = 2645.93 J

e. To get heat of reaction, we just change the sign, as it is heat released:

qreaction = -2645.93 J

f. Finally we get enthalpy of reaction dividing the heat over moles:

deltaHreaction = -2645.93 J / 0.075 moles = -35279.07 J / mol = -35.28 kJ/mol

Thermochemical equation refers to the chemical rection and its heat of solution:

NaOH + HCl -> H2O + NaCl     deltaHo =   -35.28 kJ

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