what is the ph of a solution prepared by dissolving 0.50 mol of acetic acid and 0.2 mol of sodium acetate in water and adjusting the volume of 0.5L if the pka for acetic acid is 4.75. of 0.010 mol of sodium hydroxide is added to the buffer solution from part (a), determine the ph of the new solution
Ans. #Step 1:
Concentration of acetic acid, [AH] = Moles / Vol. in liters
= 0.50 mol / 0.5 L = 1.00 M
Concentration of sodium acetate, [A-] = 0.20 mol / 0.5 L = 0.40 M
#Step 2: Titration:
what is the ph of a solution prepared by dissolving 0.50 mol of acetic acid and...
a buffer solution of pH =5.30 can be prepared by dissolving acetic acid and sodium acetate in water. How many moles of sodium acetate must be added to 1 L of 0.25 M acetic acid to prepare the buffer? Ka(CH3COOH)=1.8 x 10^-5
A 1.00 L buffer solution with pH = 4.74 is composed of 0.30 mol acetic acid and 0.30 mol sodium acetate. A) Determine the pKa of acetic acid B) If 0.030 mol of NaOH is added, determine the pH of the solution
i need help with 6 b) please 6. a) Calculate the pH of a solution prepared by dissolving 0.0775 mol acetic acid and 0.0460 mol sodium acetate in 1 L of water. (Ka = 1.8 x 10-5) ANSWER: 4.52 b) Calculate the pH of the above solution if 0.0100 mol of KOH is added to this solution. ANSWER: 4.66
Consider a buffer solution produced by dissolving 0.260 mol of acetic acid and 0.260 mol of sodium acetate in enough water to make a 500.00 mL solution. What volume in mL of a 0.150 M solution of HCl(aq) would need to be added to this solution to make it completely ineffective as a buffer? Explain.
=Assume you have prepared 100.0 mL of a buffer solution using 0.400 mol of acetic acid (pKa = 4.74) and 0.400 mol of sodium acetate. The pH of this buffer solution is initially 4.74. After preparing this buffer solution, you added 55.0 mL of a 1.10 M NaOH solution to your buffer to see what would happen. What will the pH of this new solution be?
2) A buffer solution is prepared by dissolving Sodium Acetate and Acetic acid solutions. If the overall concentration of the solution is 0.2 M and the Ka for acetic acid is 1.74 x 105 (A) What is the buffer ratio? (B) What are the individual concentrations of Acetic acid and sodium acetate needed to prepare the buffer? PH = 5 Can you please write down any assumptions needed for this particular problem? My Professor needs to see thought process. Thank...
A solution is prepared by dissolving 0.23 mol of formic acid and 0.27 mol of sodium formate in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of NaOH to this buffer solution causes the pH to increase slightly. The pH does not increase drastically because the NaOH reacts with the present in the buffer solution. The K, of formic acid is 1.8 x 10+ A formic acid B. sodium formate C water D. sodium E....
a solution is prepared by mixing 2.50 g of acetic acid (CH3CO2H, FW=60g/mol, Ka=1.75x10^(-5) with 4.70 g of sodium acetate (CH3CO2Na, FW=82g/mol) and adding water to a 500 mL volume. Note that sodium acetate yields Na+ and CH3COO-, the conjugate base of acetic acid. a.) what is the pH? b.) 15mL of 0.50 M HCl were added to the 500 mL solution, what is the pH after addituon of acid? write answer to 3 sig. figured.
a)Calculate the change in pH that occurs when 0.00100 mol of gaseous HCl is added to a buffer solution that is prepared by dissolving 4.92g of sodium acetate(molar mass = 82.03gmol^-1) in 250 ml of 0.150 mol L^-1 of acetic acid solution? Assume no change in volume upon addition of either the sodium acetate or HCl. (Ka for acetic acid = 1.8*10^-5) b)Calculate the change in pH that occurs when 0.00100 mol of gaseous HCl is added to a buffer...
3. One liter of buffer solution was prepared by mixing 0.1 mole of acetic acid CH3COOH and 0.05 mole of sodium acetate CH3COONa. Calculate a. pH of that solution b. How much of a strong base, say NaOH, in mol/L needs to be added to that solution to change its pH to 6.0? Notes and useful data: For acetic acid pK4.75 For carbonic acid pKa 6.3 and pKa 10.3 Sodium acetate CH3COONa dissociates entirely to Na'CH3COO