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- What is the pH of a solution of a freshwater solution at 25C containing 3mM DIC = H2CO3 + HCO3- + CO3^2? ( Please type answer note write by hend )The weak acid HXO2 has a dissociation constant of Ka = 4.50 × 10-10. A lab worker prepares of 1.500 M solution of this weak acid. What is the pH of this solution? Provide answer to 3 significant figures and no scientific notation please.Only typed explanation otherwise leave it The acid ionization constants of sulfurous acid (H2SO3) are Ka1 = 1.7 × 10−2 and Ka2 = 6.4 × 10−8 at 25.0 °C. Calculate the molar concentration of hydrogen sulfite ion of a 0.163 M aqueous solution of sulfurous acid. ANS: 0.045 Can you please show me how you get this answer!
- 4. Calculate the pH of a 0.340 M NaO₂CCO₂H solution. pK₁ = 1.250 and pK2 = 4.266. Please type answer not write by hendGive typed explanation not written Consider the following Ka values for phosphoric acid, H3PO4.Ka1 = 7.1 x 10-3Ka2 = 6.3 x 10-8Ka3 = 4.2 x 10-13a.) What is the pH of a 4.0 M solution of H3PO4? b.) Determine the effectiveness of H2PO4acting as a base by calculating its Kb. c.) What is the pH of a 4.0 M solution of Na3PO4 (not H3PO4)?For H2S, Ka1 = 9.6 10-8, and since Ka2 = 1.3 10-14, 0.04 M Na2S Calculate the pH of the solution by systematic method
- Write the balanced ionization or dissociation chemical equation of HC6H5CO2 in water. Include all phases. HC6H5CO2 SubmitRequest Answer Part B Determine the pH of a 0.461 M HC6H5CO2 M solution if the Ka of HC6H5CO2 is 6.5 x 10-5. 9.48 2.26 5.48 4.52 11.74How to answer for Ka/ Kb and calculated pH of a solution?I think solution 1,2 only has one Ka/Kb value then the other Sol. 3,4. Then with calculated pH, pH = -log[H+] (acid) and pH = 14 + log[OH-] (base) is utilized. Though I do not know how getting the Ka/Kb will be used to get pH Given: In a 10.0 mL 0.10 M CH3COOH solution, a 15 mL 1.00 M HCl solution was added. Compute for it's ka/kb and pH.The pH of 8.35 is wrong, but I'm not sure how to get the right answer
- A solution is made by dissolving 43.5 g of Ba(NO₂)2 in 500.0 mL of water. (a).As NO₂ is a base, write the basic equilibrium equation that exists in solution. (b).What is the value of Kb for NO₂? The Ka of HNO₂ is 4.5 × 10^-4 (c).Determine the pH of the solution Attempts all parts otherwise i will downvote....please provide handwritten solution ASAP.Choose the correct answer from among the alternatives; To make the sentence true: the pH of a solution of oxalic acid (H2C2O4) with a concentration of 0.150M at 298K is 1.16, so the Ka value for the acid ---------------------- A)10 - 2 x 9.5 B)10 - 12 x 5.9 C) 10-12 x 9.5 D) 2-10 x 5.9What is the pH of a solution of 0.25M K3PO4, potassium phosphate? Given Ka1 = 7.5*10^-3Ka2 = 6.2*10^-8Ka3 = 4.2*10^-13 I know there is another post here with the same question but nobody explained anything. Where does the K3 go? Why does everyone I see solve this just ignore it and go to H3PO4? Please type answer note write by hend