The solubility of SrSO_4 in water is 104 mg L^-1. The solubility product for SrS

Question

The solubility of SrSO_4 in water is 104 mg L^-1. The solubility product for SrSO_4 can be calculated to be 3.20 times times 10^-7. 5.66 times times 10^-4 0.320. 19.1. For a chemical reaction, delta H degree = +21.16 kJ mol^-1. The value of delta S degree for the same reaction was +74.50 J mol^-1 K^-1. If, since the physical states do not change, the value of delta H degree and delta S degree do not change with temperature over the range of interest, what should be the value of delta G_T degree 20.12 kJ mol^-1 -40.16 kJ mol^-1 -90.00 kJ mol^-1 60.00 kJ mol^-1 Calculate the concentration of iodate ions in a saturated solution of lead (II) iodate, Pb (IO_3)_2. The K_sp = 2.6 times 10^13. 4.0 times 10^-5 M 8.0 times 10^-5 M 5.1 times 10^-5 M 3.2 times 10^-5 M 6.4 times 10^-5 M Will a precipitate form when 20.0 mL of 1.8 times 10^-3 M Pb(NO_3)_2 is added to 30.0 mL of 5.0 times 10^-4 M Na_2SO_4? The K_sp of (PbSO_4) is 6.3 times 10^-7. yes, because the ion product K_sp

Explanation / Answer

5) Given H0 = +21.16 kJ/mol and S0 = +74.50 J/mol.K; we know that H0 and S0 do not change over the given temperature range. Hence we can use the relation

G0T = H0 – T*S0 where T is the temperature in the absolute scale. Given the temperature of interest is 550C, we have T = (550 + 273) K = 823 K.

Plug in values to obtain

G0T = (+21.16 kJ/mol) – (823 K)*(+74.50 J/mol.K) = (+21.16 kJ/mol) – (61313.5 J/mol) = (+21.16 kJ/mol) – (61313.5 J/mol)*(1 kJ/1000 J) = (+21.16 kJ/mol) – (61.3135 kJ/mol) = (21.16 – 61.3135) kJ/mol = -40.1535 kJ/mol -40.16 kJ/mol (ans).

Ans: (B) -40.16 kJ mol-1

6) Write down the dissociation equation:

Pb(IO3)2 (s) <====> Pb2+ (aq) + 2 IO3- (aq)

Let the molar solubility of Pb2+ be s mol/L; therefore, as per the stoichiometry of the reaction, the solubility of IO3- will be 2s mol/L.

Now, Ksp = [Pb2+][IO3-]2 = (s).(2s)2

===> 2.6*10-13 = 4s3

===> s3 = 6.5*10-14

===> s = 4.021*10-5

Therefore, [Pb2+] = 4.021*10-5 mol/L and

[IO3-] = 2*(4.021*10-5 mol/L) = 8.042*10-5 mol/L

The closest answer is 8.0*10-5 M (mol/L = M); hence (B)

Ans: (B) 8.0*10-5 M

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