Chemical reaction (dissociation) 1: C₂O₄H₂(aq) ⇄ C₂O₄H⁻(aq) + H⁺(aq).
Chemical reaction (dissociation) 2: C₂O₄H⁻(aq) ⇄ C₂O₄²⁻(aq) + H⁺(aq).
c(C₂O₄H⁻) = c(H⁺) = x.
c(C₂O₄H₂) = 0.0269 M.
pKa₁ = 1.23.
Ka₁ = 10∧(-1.23) = 0.059.
Ka₁ = c(C₂O₄H⁻) · c(H⁺) / c(C₂O₄H₂).
0.059 = x² / (0.0269 M - x).
Solve quadratic eqaution: x = c(H⁺) = 0.02 M.
pH = -log(0.02 M) = 1.7.
There are 2 significant figures.
"Rule 3: Trailing zeros are significant if a decimal point is shown in the number, but may or may not be significant if no decimal point is shown."
Answer:
M = 16.8 M
Explanation:
<u>Data:</u> HNO3
moles = 12.6 moles
solution volume = 0.75 L
Molarity is represented by the letter M and is defined as the amount of solute expressed in moles per liter of solution.
The data is replaced in the given equation:
Answer:
0.10M HCN < 0.10 M HClO < 0.10 M HNO₂ < 0.10 M HNO₃
Explanation:
We are comparing acids with the same concentration. So what we have to do first is to determine if we have any strong acid and for the rest ( weak acids ) compare them by their Ka´s ( look for them in reference tables ) since we know the larger the Ka, the more Hydronium concentration will be in these solutions at the same concentration.
HNO₃ is a strong acid and will have the largest hydronium concentration.
HCN Ka = 6.2 x 10⁻¹⁰
HNO₂ Ka = 4.0 x 10⁻⁴
HClO Ka = 3.0 x 10⁻⁸
The ranking from smallest to largest hydronium concentration will then be:
0.10M HCN < 0.10 M HClO < 0.10 M HNO₂ < 0.10 M HNO₃
I know CL is chlorine and NA is sodium.
