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- The pH of 30.0 mL of 0.80 M Benzoic acid (Ka = 5.6 x 10-5) is _____________ (with 3 significant figures)Pls help Calculate the pH of a solution (aq., 25 oC) prepared by dissolving 23.632 g of solid methylammonium chloride (CH3NH3Cl) in 1.00 L of 1.10 M methylamine (CH3NH2). The Kb for methylamine is 4.40 × 10-4. Report to two decimal places. MM CH3NH3Cl = 67.518 g/molKa1= 9.6 10-8 for H2S and Ka2 = 1.3 10-14, calculate the pH of the 0.04 M Na2S solution by systematic method.
- Hydroxylamine, HONH2, readily forms salts such as hydroxylamine hydrochloride which are used as antioxidants in soaps. Hydroxylamine has Kb of 9.1 × 10–9. What is the pH of a 0.016 M HONH2 solution? Please report 2 decimal places.ICE tables for 0.10 M HC2H3O 2 + 0.010 M + 0.0010 M. (ph for 0.10 M is 3.3. pH for 0.010 M is 3.7, ph for 0.0010 M is 4.4) HA(aq)<=> H+(aq) + A-(aq) Only typed solutionCalculate the pH of a mixture containing 0.23 M HONH2 and 0.44 M HONH3Cl. (Kb = 1.1 × 10–8) What information do we need to calculate the pH? (Choose all letters that apply.) a. 0.23 M HONH2b. 0.44 M HONH3Clc. the major species in the solutiond. pOH = –log[OH–] and 14.00 = pOH + pHe. Kb = 1.1 × 10–8
- This is a dilution lab of acetic acid. There is 1.0M acetic acid with ph of 2.297, 0.1M acetic acid with ph of 2.69897, and 0.01M acetic acid with ph of 3.69897. They were all filled up to 50ml in a volumetric flask. How do you build the ICE table for the different concentrations?I have attached some images for extra information if needed.Calculate the pH of the mixture of 94 mL butanoic acid (CH3CH2CH2COOH) 0.189 mol/L with 109 mL sodium butanoate (CH3CH2CH2COONa) 0.207 mol/L.Give an answer with at least 3 decimal places.Calculate the pH of 0.116 M NaC2H3O2 (sodium acetate) C2H3O2-(aq) + H2O(I) <-> HC2H302(aq) + OH-(aq) Ka for HC2H3O2 = 1.75 x 10^- 5 Only typed solution