A mixture of 10 moles of A, 5 moles of B and 15 moles of C is placed in a one liter container at a constant temperature. The rea
1 answer:
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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
Answer:
grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions
Explanation:
We will first write the balanced equation for this scenario
3 CaCl2 + 2 Na3PO4 ----> 6 NaCl + Ca3 (PO4)2
3 Mg(NO3)2 + 2 Na3PO4 -----> 6 NaNO3 + Mg3 (PO4)2
The ratio here for both calcium chloride and magnesium nitrate is
The number of moles of each compound is equal to
Using the mole ratio of 3:2, convert each to moles of sodium phosphate.
mole of CaCl2 is equal to
Na3PO4
mole of CaCl2 is equal to
Na3PO4
Converting moles of sodium phosphate to grams of sodium phosphate we get
g/mol
grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions