Web⇒ −pH = − log (4.8 × 10 4) = 3.32 OR pH = pK a + log [F ]− [HF] = − log (7.2 × 10−4) + log 0.15 0.10 M M = 3.14 + 0.18 = 3.32 One point is earned for indicating that the resulting solution … WebAug 28, 2013 · In this study, we prepared hematite photoanodes hydrothermally from precursor solutions of 0.1 M FeCl3 at pH 1.55 with a background electrolyte of 1.0 M …
Ka For Hf - BRAINGITH
WebHF or HBr B. NaF or HF C. CH4 or C4H10 D. NF3, HF, F2 E. He, CH4, NH3 F. H2O, SO2, 02. Answer: a. HF has the highest boiling point among the two substances because of hydrogen bonding a very strong dipole-dipole interaction. b. Explanation: ... What is the pH of a 2.0 M NaF solution? Ka of HF is 7.2 x 10-4a. 5.28b. 8.66c. 4.57d. 8.72 ... WebJan 30, 2024 · Plugging these new values into Henderson-Hasselbalch gives: pH = pK a + log (base/acid) = 3.18 + log (0.056 moles F - /0.11 moles HF) = 2.89. Thus, our buffer did what it should - it resisted the change in pH, dropping only from 3.00 to 2.89 with the addition of … The Henderson-Hasselbalch approximation allows us one method to approximat… mary greenhow
Calculate the pH of a 0.80 M aqueous solution of NaF (K_a for HF …
WebMar 4, 2024 · Since NaF is the conjugate base of HF, we can use the stoichiometry of the acid-base reaction to find that: [A-]/ [HA] = [NaF]/ [HF] = 0.200/0.300 = 0.667 Now we can plug in the values into the Henderson - Hasselbalch equation: pH = pKa + log ( [A-]/ [HA]) pH = -log (6.8 × 10⁻⁴) + log (0.667) pH = 3.17 + (-0.177) pH = 2.99 WebNov 28, 2024 · The Henderson-Hasselbalch equation is used to estimate the p H of a buffer. Expert Answer Now, using the Henderson-Hasselbalch equation: p H = p K a + log [ F] [ H F] p H = p K a + log [ N a F] [ H F] p H − p K a = log [ N a F] [ H F] log ( 10 ( p H − p K a)) = log [ N a F] [ H F] Applying anti-log on both sides, we get: hurricane 2004 fl