Answer:
pH = 12.8
Explanation:
HF + NaOH → F⁻ + Na⁺ + H₂O
<em>1 mole of HF reacts with 1 mole of NaOH</em>
<em />
Initial moles of HF and NaOH are:
HF = 0.018L × (0.308mol / L) = 5.544x10⁻³mol HF
NaOH = 0.023L × (0.361mol / L) = 8.303x10⁻³mol NaOH
That means moles of NaOH remains after reaction are:
8.303x10⁻³mol - 5.544x10⁻³mol = <em>2.759x10⁻³moles NaOH</em>
Total volume is 18.0mL + 23.0mL = 41.0mL = 0.0410L
Molar concentration of NaOH is
2.759x10⁻³moles NaOH / 0.0410L = 0.0673M = [OH⁻]
pOH = - log [OH⁻] = 1.17
As pH = 14 - pOH
<em>pH = 12.8</em>
<em></em>
When the Ph can be determined by the dissociated H+ of the strong acid, and
we have here HNO3 is the strong acid so, it will ionize completely to H+ & NO3-
HNO3 ↔ H+ + NO3-
∴ [H+] = [NO3-] = 0.075 m
∴PH = - ㏒ [H+]
= - ㏒ 0.075
= 1.12
Based on that, alpha-particle get attracted and many of the particles were not gone away without deflection
Answer:
H₃O⁺/OH⁻
Explanation:
Conjugate acid-base pair are defined as compounds which differ by H⁺. The acid is the compound that contains the additional H⁺
The conjugate base of H₂SO₃ is HSO₃⁻. The conjugate base of NH₄⁺ is NH₃. The conjugate acid of C₂H₃O₂⁻ is HC₂H₃O₂ and conjugate base of H₃O⁺ is H₂O, not OH⁻. Thus, right answer is <em>H₃O⁺/OH⁻</em>.
I hope it helps!