H3c6h5o7(aq) + 3nahco3(aq) → 3co2(g) + 3h2o(l) + na3c6h5o7(aq) calculate the number of grams of baking soda (nahco3; molar mass
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Answer:
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- <u>- 0.380ºC</u>
Explanation:
The lowering of the freezing point of a solvent is a colligative property ruled by the formula:
Where:
- ΔTf is the lowering of the freezing point
- Kf is the molal freezing constant of the solvent: 1.86 °C/m
- m is the molality of the solution
- i is the van't Hoff factor: the number of particles (ions) per unit of ionic compound.
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<u>a) molality, m</u>
- m = number of moles of solute/ kg of solvent
- number of moles of CaI₂ = mass in grams/ molar mass
- number of moles of CaI₂ = 25.00g / 293.887 g/mol = 0.0850667mol
- m = 0.0850667mol/1.25 kg = 0.068053m
<u>b) i</u>
- Each unit of CaI₂, ideally, dissociates into 1 Ca⁺ ion and 2 I⁻ ions. Thus, i = 1 + 2 = 3
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<u>c) Freezing point lowering</u>
- ΔTf = 1.86 °C/m × 0.068053m × 3 = 0.3797ºC ≈ 0.380ºC
Answer:
1. pH = 1.23.
2.
Explanation:
Hello!
1. In this case, for the ionization of H2C2O4, we can write:
It means, that if it is forming a buffer solution with its conjugate base in the form of KHC2O4, we can compute the pH based on the Henderson-Hasselbach equation:
Whereas the pKa is:
The concentration of the base is 0.347 M and the concentration of the acid is 0.347 M as well, as seen on the statement; thus, the pH is:
2. Now, since the addition of KOH directly consumes 0.070 moles of acid, we can compute the remaining moles as follows:
It means that the acid remains in excess yet more base is yielded due to the effect of the OH ions provided by the KOH; therefore, the undergone chemical reaction is:
Which is also shown in net ionic notation.
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