I have a buffer solution that consists of 0.200M benzoic acid, and 0.220M sodium benzoate. If the Ka (benzoic acid) =6.46(10^-5), what is the pH?
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I have a buffer solution that consists of 0.200M benzoic acid, and 0.220M sodium benzoate. If the Ka (benzoic acid) =6.46(10^-5), what is the pH?
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- Estimate the pH that results when the following two solutions are mixed. a) 50 mL of 0.3 M CH3COOH and 50 mL of 0.4 M KOH b) 100 mL of 0.3 M CH3COOH and 50 mL of 0.4 M NaOH c) 150 mL of 0.3 M CH3COOH and 100 mL of 0.3 M Ba(OH)2 d) 200 mL of 0.3 M CH3COOH and 100 mL of 0.3 M Ba(OH)2What is the pH of a buffer that is 0.150 M in a weak acid and 0.150 M in the acids conjugate base? The acids ionization constant is 6.8 106.8-71 Explain why you do not need to know the chemical formula of a buffer compound to use it.
- A buffer is made by adding 0.18 mol/L of sodium propanoate (NaPr) to 0.12 mol/L solution of propanoic acid (HPr, K«= 1.3 x 105). Assume the volume of the solution to be 1.0 L. a) Write the chemical equation for the acid hydrolysis reaction. | b) What is the acid-base conjugate pair that governs the pH of the buffer? c) Calculate the pH of the buffer. Be sure to justify and validate any approximations used.You are studying a clear solution and you added the pH indicator methyl violet. The colour range of the indicator methyl violet in a clear solution when changing from acidic to basic is yellow (pH 0) to blue purple (pH 1) to violet (pH 2). You initial pH of the solution when tested with a pH meter is O.2. You are going to add 250 drops of 0.1 M HCI. Please select the most appropriate answers to the following two questions. What is the initial colour of the solution at pH 0.2? What is the colour of the solution and what will the pH be after the addition of the HCI? Select 2 correct answer(s) The colour of the solution after the addition of HCI will be clear and the pH will be less than 0.2. The colour of the solution after the addition of HCI will be violet and the pH will be higher than 0.2.An organic acid (HA) has a molecular weight of 100g/mol, a Kow = 5.6 and a K, = 2.7 x 10-2. If originally 2.0 g of the acid is dissolved in 100 mL of octanol (there is no dissociation in octanol), which is then placed in contact with 100 mL of water, what will be the pH of the water? (Consider the equilibrium processes to be sequential and unrelated chemically) Possible answers: 1.89, 1.48, 2.01, 1.74, 1.61
- A solution is prepared that is initially 0.45M in pyridine (CÂHÂN), a weak base, and 0.50M in pyridinium bromide (C5H₁NHBr). Complete the reaction table below, so that you could use it to calculate the pH of this solution. Use x to stand for the unknown change in [où¯]. You can leave out the symbol for molarity. initial change final [C,H,N] [C₂H₂NH*] [OH-] 0 0 0 0 DO X ArYou are given 250.0 mL of 0.100 M CH3CH2COOH (propionic acid, Ka = 1.35 × 10−5 ). You wish to adjust the pH by adding an appropriate solution. What volume would you add of a) 1.00 M HCl to lower the pH to 1.00; b) 1.00 M CH3CH2COONa to raise the pH to 4.00?Given that Ka’s for hydrofluoric acid (HF) and boric acid (H3BO3) are 6.3 × 10^–4and 5.4 × 10^–10, respectively, calculate the pH of the following solutions: (a)The mixture from adding 50 mL 0.2 M HF to 50 mL 0.5 M sodium borate (NaH2BO3). (b)The mixture from adding an additional 150 mL 0.2 M HF to the solution in (a), i.e., a total of 200 mL 0.2 M HF was added to 50 mL 0.5 M NaH2BO3.
- What is the pH of a buffer system prepared by dissolving 12.00 grams of NH4Cl and 40.00mL of 12 M NH3 in enough water to make 1.0L of solution? Kb=1.80*10^-5 for NH3- what is the pH of a solution made by mixing 30 mL of 0.100 M HCl with 40.0 mL of 0.100 M KOH? Assume that the total volume is 70mLAn organic acid(HA) has a molecular weight of 100. g mol^-1, a Kow=5.6 and a Ka=2.7x10^-2. If originally 2.0g of the acid is dissolved in 100mL of octanol (there is no dissociation in octanol), which is then placed in contact with 100mL of water, what will be the pH of the water? (consider the equilibrium processes to be sequential and unrelated chemically)(a) Calculate the pH in the solution formed by adding 10.0 mL of 0.050 M NaOH to 40.0 mL of 0.0250 M benzoic acid (C6H5COOH, Ka = 6.3 * 10-5). (b) Calculate the pH in the solution formed by adding 10.0 mL of 0.100 M HCl to20.0 mL of 0.100 M NH3.