What is the pH at the equivalence point when 100.00 mL of a 0.175 M solution of

Question

What is the pH at the equivalence point when 100.00 mL of a 0.175 M solution of acetic acid (CH_3 COOH) is titrated with 0.10 M NaOH to its end point? Analyze For CH_3 COOH, the K_a = 1.76 times 10^-5. To calculate the pH of the solution at the equivalence point, we need to recognize that at that point all of the CH_3 COOH is converted to CH_3 COO^-. We use K_b, therefore, for the ICE table calculation. We also need to take into account the added volume of the solution when NaOH titrant is added. CH_3 COO^- (aq) + H_2 O (l) CH_3 COOH (aq) + OH^- (aq)

Explanation / Answer

Acetic acid = 0.100 L x 0.175 M = 0.0175 M
NaOH required = 0.0175
volume NaOH = 0.0175 / 0.10 M = 0.175 L
total volume = 0.100 + 0.175 = 0.275 L
Acetate formed = 0.0175 / 0.275 =0.0636 M

the equilibrium is
CH3COO- + H2O CH3COOH + OH-
Kb = Kw/Ka

= 1.0 x 10^-14 / 1.8 x 10^-5

= 5.6 x 10^-10

= x^2 / 0.0636-x
x = [OH-]= 6.0 x 10^-6 M
pOH = 5.2
pH = 14 - 5.2 = 8.8

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