As dissociation of acetic acid will give ,CH3COO- and H+
Hence its pH = - log (H+)
pH = - log (1.49 x 10 -3) = - log(1.49) - log 10 -3 = -0.1731 + 3 = 2.82
*For y = log (a ±S(a)), uncertainty in y i.e .S(y) = 0.434*(S(a)/a)
Hence,
uncertainty in pH will be 0.434 *(1.49 x 10 -3 /4 x 10 -5) = .01165
Hence,
pH of solution with uncertainty = 2.82 ± 0.01165
If a solution of acetic acid (K = 1.8 x 10-5) has a pH of 2.90, calculate the original (initial) concentration of acetic acid (HC,H,O) (Report your answer in 1 sig. fig. Example: .03942 would be reported as 0.04) Be sure to include a zero in front of the decimal point. QUESTION 37 5 points Save Answer The K, for benzoic acid C.H.COOH is 6.3 x 106. Calculate the equilibrium concentrations of H,0* in the solution if the initial concentration...
Calculate the pH of a .30 M solution of acetic acid (CH3COOH, Ka= 1.8 x 10^-5) at 25 degrees Celsius.
a) The degree of ionization of acetic acid (CH3COOH) in a 0.1 M aqueous solution at 25 oC is 0.013. Ka 1.7 X 10-5. What is the ph of this solution? What is the deg of ionization b) A solution is prepared to be 0.1 M acetic acid CH,COOH and 0.2 M CHCOONa. What is the pH of this solution at 25°C ? K, for acetic acid at this temperature -1.7 X10
3. Show how acetic acid, CH3COOH, ionizes in solution and calculate the pH of a 0.30 M solution of the acid. Ką for acetic acid = 1.8 x 10-5. 4. Show how aniline, CH5NH2, ionizes in solution and calculate the pH of a 0.25 M solution of the compound. Kb = 3.8 x 10-10
What is the pH of a 0.25 M solution of acetic acid? The Ka for acetic acid is 1.74 x 10-5 M. Round the answer to one decimal place.
pH of solution of acetic acid (K, - 1.8 10-5) is 3. What is concentration of the acid? 0.018 M 0.056 M 0.006 M O 1.8 M
Calculate the [H+] and pH of a 0.0035 M acetic acid solution. The Ka of acetic acid is 1.76 x 10-5. Use the method of successive approximations in your calculations. [H+] = _______ MPH = _______
A 1.49 L buffer solution consists of 0.286 M propanoic acid and 0.189 M sodium propanoate. Calculate the pH of the solution following the addition of 0.066 mol HCL. Assume that any contribution of the HCl to the volume of the solution is negligible. The K, of propanoic acid is 1.34 x 10-5.
starting with a solution at equilibrium containing .5 M acetate and .5 M acetic acid, then adding .1 M sodiun acetate, what would be the pH of the original solution and the solution after adding sodium acetate. the dissociation constant (Ka) for acetic acid is 1*10^-6
The Ka value for acetic acid, CH3COOH(aq) , is 1.8×10-5 M . Calculate the pH of a 2.80 M acetic acid solution.Calculate the pH of the resulting solution when 2.50 mL of the 2.80 M acetic acid is diluted to make a 250.0 mL solution.