WebMar 5, 2024 · Calculate the pH of a 0.39 M CH3COONa solution. (Ka for acetic acid = 1.8 × 10−5.) ... 5.0 (387) See more tutors. find an online tutor. Chemistry tutors; Physical Chemistry tutors; Organic Chemistry tutors; AP Chemistry tutors; Biochemistry tutors; Thermodynamics tutors; Chemical Engineering tutors; Heat Transfer tutors; WebQuestion: Calculate the pH of a 2.0 L buffer solution which is 1.2 M CH3COONa and 0.8 M CH3COOH. Kb (CH3CO2-) = 5.56E-10. Calculate the pH of a 2.0 L buffer solution which is 1.2 M CH3COONa and 0.8 M CH3COOH. Kb (CH3CO2-) = 5.56E-10. Expert Answer. Who are the experts? Experts are tested by Chegg as specialists in their subject area. We ...
某同学测得物质的量浓度均为0.01mol?L -1的CH3COOH和CH3COONa …
WebMar 5, 2024 · Calculate the pH of a 0.39 M CH3COONa solution. (Ka for acetic acid = 1.8 × 10−5.) Wyzant Ask An Expert. Chemistry. Javier C. asked • 03/05/21. Calculate the pH of … WebA 2.0 L buffer solution contains 0.05 M CH3COOH and 0.04 M CH3COONa. Calculate the new pH after adding 0.02 mol of NaOH. Ka(CH3COOH) = 1.8 × 10−5. Expert Answer. Who are the experts? Experts are tested by Chegg as specialists in their subject area. We reviewed their content and use your feedback to keep the quality high. flying dnd creatures
pH of 0.1 M CH3COOH - Wolfram Alpha
Web102[OH ]2 = 0.1 × 10 9.24 and pH = 7.6. This is very close to the value obtained by the students. The CO 2 disso-lution decreases the pH of the sodium acetate solution by more than one unit. An experimental proof of this effect can be given. When pure N 2 is bubbled through the solution, the pH increases. However, this increase is very slow ... WebCH3COOH (10.0 g NaCH3COO)(1 mol/82.03 g) = 0.122 mol NaCH3COO Substitute these values, along with the Kavalue, into the above equation and solve for the hydronium ion concentration. Convert the hydronium ion concentration into pH. [H3O+] = (1.7 x 10-5)(0.200/0.122) = 2.79 x 10-5 pH = 4.56 WebJan 5, 2016 · Even without doing any calculation, you can look at this equation and say that when you have equal concentrations of weak base and conjugate acid, the pOH of the solution will be equal to pKb, since [NH+ 4] = [NH3] ⇒ log( [NH+ 4] [NH3]) = 0 Since pkb = − log(Kb) you will have pOH = − log(1.8 ⋅ 10−5) = 4.74 The pH of the buffer will thus be flying distance from south africa to dubai