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2.083 Liters of 6.0 M solution sulfuric acid is required. This solved using molecular calculations and Titration.
Solution:
Moles of hydrogen gas =
Then 12.5 moles of hydrogen will be obtained from Moles of Sulfuric acid = 12.5 mol
Molarity of the sulfuric acid solution = 6.0 M = 6 mol/ l
6M =
where V is the volume needed
V = 2.083 l
<h3>What is Titration?</h3>- Titration, commonly referred to as titrimetry, is a typical quantitative chemical analysis method used in laboratories to ascertain the unidentified quantity of an analyte .
- Titration is frequently referred to as volumetric analysis because it relies heavily on volume measurements. The titrant or titrator is a reagent that is prepared as a standard solution.
- To determine concentration, a solution of the analyte or titrand reacts with a known concentration and volume of the titrant. The titration volume is the amount of titrant that has responded.
- Titrations come in a variety of forms with various protocols and objectives. Redox and acid-base titrations are the two most typical types of qualitative titrations.
To learn more about titration with the given link
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First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2 so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄ ⇄ H⁺ + HC₂O₄⁻
0.0356 M 0 0
0.0356 - x x x
Ka1 =![\frac{[H^+][HC2O4^-]}{[H2C2O4]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5BHC2O4%5E-%5D%7D%7B%5BH2C2O4%5D%7D%20)
= x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2 so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄ ⇄ H⁺ + HC₂O₄⁻
0.0356 M 0 0
0.0356 - x x x
Ka1 =
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6