Determine the pH at the equivalence (stoichiometric) point in the titration?

Question

of 46.8 mL of 0.138 M acetic acid by 0.213 M KOH. The Ka of CH3COOH is 1.8 x 10-5.

Can someone work this our for me step by step b/c i do not understand where I am getting some of the values from.  Thanks!

jacob s · Accepted Answer

46.8mL of 0.138M CH3COOH 0.138M
moles CH3COOH = 0.138 M x 0.0488 L = .006734 moles CH3COOH
moles KOH= 0.006743
V = KOH = 0.00634/ 0.138 = 0.0488 L
Total volume = 0.976 L
[CH3COOH-] = 0.006734 / 0.0976 L = .068996M

K = 10^-14 /1.8 x 10^-5= 5.6 x 10^-9 = x^2/ .068996-x
x = [OH-] = .00002M
pOH = 4.70
pH=9.30

Anonym · Answer

pH = pKa + log [salt] / [acid] pKa=4.seventy 4 concentation of acetic acid in litre=0.0.5 => concentration of acid in 50 ml=0.0.5*50/one thousand =7.5*10^-4 concentration of NaOH=0.03 pH=4.seventy 4+log[7.5*10^-4} /[0.03] pH=4.seventy 4-one million.602 pH=30138

Determine the pH at the equivalence (stoichiometric) point in the titration?

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