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 =
= x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6
Answer:
En el caso del sodio, la valencia es 1, ya que tiene un solo electrón de valencia, si pierde un electrón se queda con el último nivel completo.
Explanation:
Grupo de la tabla periódica Electrones de valencia
Grupo 14 (IV) (Grupo del carbono) 4
Grupo 15 (V) (Grupo del nitrógeno ) 5
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Answer:
0.5mol/L
Explanation:
First, let us calculate the number of mole NaOH = 23 + 16 + 1 = 40g/mol
Mass of NaOH from the question = 30g
Number of mole = Mass /Molar Mass
Number of mole = 30/40 = 0.75mol
Volume = 1.5L
Active mass = mole/Volume
Active mass = 0.75mol/1.5L
Active mass = 0.5mol/L