The Ka value for acetic acid, CH3COOH(aq), is 1.8× 10–5. Calculate the pH of a 2

Question

The Ka value for acetic acid, CH3COOH(aq), is 1.8× 10–5. Calculate the pH of a 2.20 M acetic acid solution.

Res o 4/8/2016 11:55 PM 0 22/504/8/2016 07:45 PM 4/8/2016 11:55 PM A 22/50 4/8/2016 07:45 PM Calculator-Periodic Table Gradebook Question 15 of 20 Map neral Chemistr Donald McQuarrie Peter A. Rock Ethan Gallogly University Science Books presented by Sapling Learning The Ka value for acetic acid, CHsCOOH(aq), is 1.8x 10-5. Calculate the pH of a 2.20 M acetic acid solution. Number Calculate the pH of the resulting solution when 3.50 mL of the 2.20 M acetic acid is diluted to make a 250.0 mL solution. Number Check Answer 0 Next Check Answer Next Exit- Previous Exit Hint

Explanation / Answer

CH3COOH ---> CH3COO- + H+
2.2 0 0 (initial)
2.2-x x x (at equilibrium)

Ka = [CH3COO-][H+]/[CH3COOH]
1.8*10^-5 = x*x/(2.2-x)

since Ka is very small, x will be small and it cna be ignored as compared to 2.2
ABove expression now becomes
1.8*10^-5 = x*x/(2.2)
x = 6.29*10^-3 M

So,
[H+] = 6.29*10^-3 M

pH = -log [H+]
pH = -log (6.29*10^-3)
pH = 2.2
Answer: 2.2

--------------------------------------------
use dilution formula to final final concentration of CH3COOH
M1*V1 = M2*V2
2.2*3.5 = M*250
M = 0.0308 M

CH3COOH ---> CH3COO- + H+
0.0308 0 0 (initial)
0.0308-x x x (at equilibrium)

Ka = [CH3COO-][H+]/[CH3COOH]
1.8*10^-5 = x*x/(0.0308-x)

5.544*10^-7 - 1.8*10^-5 x = x^2
x^2 + 1.8*10^-5 x - 5.544*10^-7 = 0

solving above quadratic equation we get,
x = 0.00074 M or -0.00075 M
x can't eb negative,
so,
x = 0.00074 M

So,
[H+] = 60.00074 M

pH = -log [H+]
pH = -log (0.00074 )
pH = 3.13
Answer: 3.13

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