55) A 25.0 mL sample of an acetic acid solution is titrated with a 0.18 M NaOH s

Question

55) A 25.0 mL sample of an acetic acid solution is titrated with a 0.18 M NaOH solution. The equivalence point

is reached when 37.0 mL of the base is added. The concentration of acetic acid in the sample was _______ M.

A) 0.119 B) 1.83 x 10-4 C) 0.266 D) 0.365 E) 0.175

56) What is the pH of the resulting solution in the previous problem? Ka acetic acid = 1.8 x 10-5

A) 2.57 B) 5.72 C) 8.80 D) 10.65 E) 12.91

I know the answer (bold one) but I dont know how to get to the answer

Could you explain with detail equation?

Explanation / Answer

(55) Option (C) is the correct answer.

Explanation:

Here, you need to do for this one is to find the number of mole of OH you added and that will be the smae as the number of moles of acetic acid there was and then divide that by the 0.025 L to get thefinal M concentration.

0.18 * 0.037 = 6.66 x 10^-3 mol

So, 6.66 x 10^-3 mol / 0.025 = 0.266 M

(56) Option (A) is the correct answer.

Explanation:

Write the expression -

Ka = [H+][A-]/ [HA]-x
CH3COOH(aq) --> CH3COO- + H+
[H+] = [A-]
(-x) is excluded as Ka & conc. differs >factor 100
Ka = x^2/ [HA]

put the values -
1.8 x 10^-5 = x^2 / 0.266
=> 2.19 x 10^-3 = x = [H+]
pH = -log[H+] = 2.60

You have mentioned the correct answer as option (C) but from the above calculations, it comes as 2.60. Please check your answer.

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