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- Calculate the pH of a 0.020 M solution of phenylacetic acid, C6H5CH2COOH. What will be the pH if the solution is made 0.050 M with its sodium salt, C6H5CH2COONa? Ka= 4.9 x 10-5.Calculate the pH of a 0.295 M solution of ethylenediamine (H₂NCH,CH,NH₂). The pKa values for the acidic form of ethylenediamine (HNCH,CH,NH) are 6.848 (pKa1) and 9.928 (pKa2). pH = Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. [H2NCH2CH2NH₂] = [H2NCH,CH,NH] = [HNCH,CH,NH3] = M M MCalculate the pH of a 0.025 M solution of propanoic acid (Ka = 1.3 x 10-5). The answer is 1.6
- K Calculate the pH of a 0.295 M solution of ethylenediamine (H₂NCH₂CH₂NH₂). The pKa values for the acidic form of ethylenediamine (HFNCH₂CH₂NH3) are 6.848 (pKat) and 9.928 (pK₁2). pH = Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. [H₂NCH₂CH₂NH₂] = [H₂NCH₂CH₂NH] = [HNCH₂CH₂NH3] = M M MThe Ka of CH3COOH is 1.8 x 10-5. Calculate the pH of a solution containing both 0.20 M CH3COOH and 0.30 M CH3COONa.The amount of tartaric acid is responsible for the tartness of wine and controls the acidity of the wine. Tartaric acid also plays a very significant role in the overall taste, feel and color of a wine. Tartaric acid is a diprotic organic acid The chemical formula for tartaric acid is C4H6O6 and its structural formula is HO2CCH(OH)CH(OH)CO2H. A 50.00 mL sample of a white dinner wine required 21.48 mL of 0.03776 M NaOH to achieve a faint pink color. Express the acidity of the wine in terms of grams of tartaric acid, H2C4H4O6 (M. M. = 150.10) per 100 mL of wine. Assume that the two acidic hydrogens are titrated at the end point. MM H2C4H4O6 = 150.10 MM NaOH = 40.00 Below is the balanced chemical equation for this titration.
- Calculate the pH of a 0.0015 M solution of C6H5NH2 (Kb for C6H5NH2 = 3.80 ⨉ 10-10).What is the pH of an aqueous solution containing 0.50 mol L–1 ethanoic acid, 0.30 mol L–1 sodium ethanoate, and 0.10 mol L–1 HCl? [The pKa of ethanoic acid is 4.75.]Calculate the pH of a 0.123 M solution of ethylenediamine (H₂NCH₂CH₂NH₂). The pKą values for the acidic form of ethylenediamine (H3NCH₂CH₂NH³) are 6.848 (pKal) and 9.928 (pKa2). pH = Calculate the concentration of each form of ethylenediamine in this solution at equilibrium. [H₂NCH₂CH₂NH₂] : = [H₂NCH₂CH₂NH3] = M M
- Calculate the pH of a solution that is 0.132 M in diethylamine, (CH3CH2)2NH, and 0.145 M in diethylammonium chloride, (CH3CH2)2NH2CI. (CH3CH2)2NH + H20 - (CH3CH2)2NH2* + OH Kp = 6.9 x 10-4 pH = 10.80 pH = 12.14 pH = 11.98 pH = 3.20 pH = 6.3 x 10-4What is the pH of 0.093 ?0.093 � butanoic acid, C3H7COOH? The Ka of butanoic acid is 1.5 x 10-5.The acid dissociation constant K, of trimethylacetic acid (HC (CH3) CO2 is 9.33 × 10 *. Calculate the pH of a 6.1M solution of trimethylacetic acid. Round your answer to 1 decimal place. PH