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Please calculate the pH of a 0.5L solution prepared with 0.05 M acetic acid (pK = 4.76) before and after the addition of 2 mL of 5 M NaOH. In a secondary calculation, please calculate the pH of this solution if the buffer was not present.
Hint: 1 Do not forget to calculate the equilibrium concentrations of acid and conjugate base as a first step.
Hint: 2 Your goal is to calculate 3 pH values
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- Determine the pH of a solution of HF by constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the pH. The Ka of HF is 6.8 x 10-4. Complete Parts 1-3 before submitting your answer. NEXT > A 0.30 M aqueous solution of HF is prepared. Fill in the ICE table with the appropriate value for each involved species to determine concentrations of all reactants and products. Initial (M) Change (M) Equilibrium (M) -2x 6.8 x 10-4-2x 0 0.30 + x 1 HF(aq) 0.30 0.30 - x + H₂O(l) 6.8 x 10-4 2 0.30 + 2x 0.15 0.30 - 2x H3O+(aq) + +X 3 6.8 x 104 + x -X 6.8 x 10-4-x F-(aq) ✓ RESET +2x 6.8 x 10-4 + 2xThe pH of a solution made from a 0.75 M acetic acid solution and a 0.54 M sodium acetate solution is (K, = 1.8 x 10-5) Your Answer: Answer Question 7 1) Listen The pH of 300 mL solution made of 1.10 M acetic acid and 0.80 M potassium acetate is (Ka = 1.8 x 105) after the addition of 0.03 moles HCI Your Answer: Answer Question 8 1) Listen The pH of 300 mL solution made of 0.97 M acetic acid and 1.19 M potassium acetate is (Ką = 1.8 x 10-5) after the addition of 0.010 moles NaOH Your Answer: Answer+ Base/Acid Ratios in Buffers Just as pH is the negative logarithm of [H3O+]. PK₁ is the negative logarithm of Ka. PK₂ = -log Ka The Henderson-Hasselbalch equation is used to calculate the pH of buffer solutions: pH=pK₁ +log base [acid] Notice that the pH of a buffer has a value close to the PK₂ of the acid, differing only by the logarithm of the concentration ratio [base]/[acid]. The Henderson-Hasselbalch equation in terms of pOH and PKb is similar. lacid] pOH = pk + logad base pH 3.74 [acetic acid] ten times greater than [acetate] Submit ✓ Correct ▼ Part B pH = 4.74 Previous Answers [acetate] = [acetic acid] mass of NH4Cl = pH = 5.74 [acetate] ten times greater than [acetic acid] Value How many grams of dry NH4Cl need to be added to 2.00 L of a 0.100 M solution of ammonia, NH3. to prepare a buffer solution that has a pH of 8.85? Kb for ammonia is 1.8 x 10-5 Express your answer with the appropriate units. ▸ View Available Hint(s) ( 2 F Templates Symbols undo relo reset keyboard…
- Determine the concentration of CH NH¸* in a buffer solution by constructing an ICE table, writing the equilibrium constant expression, and use this information to determine the concentration of CH NH+. The Kb for CH NH₂ is 3.8 × 10-10. Complete Parts 1-3 before submitting your answer. 2 1 2 3 NEXT > 2 A buffer solution contains dissolved CH NH and CH NH Cl. The initial concentration of CH NH2 is 0.50 M. The pH at equilibrium of the buffer is 4.20. Let x represent the original concentration of CH NH+ in the water. Fill in the ICE table with the appropriate value for each involved species to determine concentrations of all reactants and products. Initial (M) Change (M) Equilibrium (M) CH NH2(aq) + H₂O(I) = OH(aq) + C HẠNH, (aq) RESET 0 0.50 x 4.20 -4.20 6.3 x 10-5 -6.3 x 10-5 1.6 × 10-10 -1.6 × 10-10 3.8 × 10-10 -3.8 × 10-10 x +4.20 x-4.20 x + 6.3 x 10-5 x - 6.3 × 10-5 x + 1.6 × 10-10 x-1.6 × 10-10 x + 3.8 × 10-10 x - 3.8 × 10-10What is the pH at which a buffer composed of NH3 and NH4+ would be most effective at resisting pH change? Round your answer to 2 decimal places. Remember you can find KA and/or KB valuesYou want to use acetic acid (pka=4.75) and NaOH to prep 1 liter 0.1 M buffer solution with a pH= 5.0. Keeping in mind that conc of acetic acid + conc of acetate= 0.1 M, how many moles of acetic acid and NaOH would you need to add to 1 L of water?
- What is the pH of a buffer solution that is 0.20 M methyl amine, CH3NH2 and 0.18 M methylammonium chloride CH3NH2Cl? The Kb for CH3NH2 is 4.4x10-4. You do not need to solve the quadratic equation for this problem. Express your answer to two decimal places.acid. 2. Hypochlorous acid, HC10, is a weak acid with Ka = 3.0 x 10-8 a. Compute the pH of the mixture after 0.094 moles of Naot is added to 1 L of a 0.155 M HC10 solution. b. Consider a mixture of 0.155 moles of HC10 and 0.184 moles of KC10 in 1 L of solution. Use the Henderson-Hasselbach equation to compute the pH of this buffer. c. Compute the pH of the mixture after 0.094 moles of NaOH is added to i L of the buffer solution described in part b. d. Compute the pH of the mixture after 0.194 moles of NAOH is added to 1 Lof the buffer solution described in part b.Using the table of the weak base below, you have chosen Pyridine as your weak base in the buffer solution. You have already added enough of the conjugate acid salt to make the buffer solution concentration at 0.62 M in this salt. The desired pH of the buffer should be equal to 4.5. Values of K, for Some Common Weak Bases 車 Conjugate Acid Name Formula 1.8 x 10-5 4.38 x 10-4 5.6 x 10-4 3.8 x 10-10 Ammonia NH3 CH;NH2 CH§NH2 CH;NH2 C;H;N NH,+ Methylamine Ethylamine Aniline Pyridine CH;NH;* CH$NH;* CH;NH;* C;H;NH 1.7 x 10-9 1. Compute the pOH of the buffer solution. (Write your answer in 1 decimal place).
- Determine the pH of a buffer solution by constructing an ICE table, writing the equilibrium constant expression, and using this information to determine the pH. Complete Parts 1-3 before submitting your answer. 3 NEXT > The buffer was prepared by dissolving 20.0 g NaCH3COO into a 500.0 mL solution of 0.150 M of CH3COOH. Assume the volume of the solution does not change. Fill in the ICE table with the appropriate value for each involved species to determine concentrations of all reactants and products. 4 Initial (M) Change (M) Equilibrium (M) 1 2 CH3COOH(aq) + H₂O(l) = H3O+ (aq) + CH3COO-(aq)What is a buffer solution? How would an 0.5 M acetic acid/acetate buffer solution respond to the addition of 0.1 M KOH? Write your full answer in the textbox.You have 0.600 L of an 0.300 M acetate buffer solution (i.e. [HC;H,O2] + [C,H¿O¿¯] = 0.300 M) at pH 5.30. What will the new pH of the solution be if you add 82.0 mL of 1.000 M HCI to it. Acetic acid has a pKa of 4.74.