Answer:
[NaOH} = 0.4 M
Explanation:
In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.
(H₂SO₄, is considered strong, but the first deprotonation is weak)
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.
In the equivalence point we know mmoles of base = mmoles of acid
Let's finish the excersise with the formula
25 mL . M NaOH = 28.2 mL . 0.355M
M NaOH = (28.2 mL . 0.355M) / 25 mL → 0.400
Answer:
The correct answer is B) HOOCCH2CH2COOH(aq)
Explanation:
Both Ka1 and Ka2 are low, so the acid will dissociate only slightly into HOOCCH2CH2COO- ions and even more slightly into -OOCCH2CH2COO- ions. The concentration of hydronium ions (H₃O⁺) will be consequently low. The species that will be in the highest concentration will be HOOCCH2CH2COOH (the weak acid not dissociated).
Answer:
'See Explanation
Explanation:
Determine the [OH−] , pH, and pOH of a solution with a [H+] of 9.5×10−13 M at 25 °C.
Given [H⁺] = 9.5 x 10⁻¹³M => [H⁺][OH⁻] = 1.0 x 10⁻¹⁴ => [OH⁻] = 1.0 x 10⁻¹⁴/9.5 x 10⁻¹³ = 0.0105M
pH = -log[H⁺] = -log(9.5 x 10⁻¹³) = - (-1202) = 12.02.
pOH = -log[OH⁻] = -log(0.0105) = -(-1.98) = 1.98
Now you use the same sequence in the remaining problems.
