- points Gvsuchem 1162 14.J.1.P001 Weak Acid/Base Titration: Understanding the M

Question

- points Gvsuchem 1162 14.J.1.P001 Weak Acid/Base Titration: Understanding the Method During a titration of a weak acid with a strong base, you are slowing converting molecules of the weak acid into molecules of its conjugate base. For the hypothetical weak acid, HA we see the following: HA (aq) + OH' (aq) A. (aq) + H2O (l) Early in the titration, this is like a limiting reagent problem where the weak acid is present in excess. Each molecule of strong base converts one molecule of HA into one molecule of the conjugate base, A", so that the [HA] decreases and the [A] increases. As these two species (HA and A) return to equilibrium the pH is controlled by the common ion effect (just like in the common ion effect problems we have already worked). General Method: In the problem below you will be adding some strong base, but not enough to reach the endpoint of the titration. Use the solution stoichiometry to find out how many moles of the weak acid remain after the reaction with the strong base. Use this and the total volume to get the new weak acid and A concentrations. Then set it up like a common ion effect problem using the Ka, [HA] and [A] which was created by the strong base. 0.98M -sodium 1.60 mL of sodium hydroxide added to the weak acid hydroxide in the buret as the titrant 18.00 mL of original 0.75 M CH3COOH sample In this problem you are titrating 18.00 mL of a 0.75 M solution of CHjCOOH with a strong base. If you add 1.60 ml of a 0.98 M solution of sodium hydroxide, what is the final pt of the acid solution? The K, for CHjCOOH is 1.8x 10,

Explanation / Answer

moles of acetic acid present = molarity* volume of acetic acid in Liters, 1000ml= 1L,

hence moles of acetic acid = 0.75*18/1000 =0.0135, moles of NaOH= 0.98*1.6/1000 =0.001568

the reaction between acetic acid and sodium hydroxide is CH3COOH+ NaOH --------->CH3COONa+ H2O

Theoretical molar ratio of CH3COOH: NaOH= 1:1

Actual molar ratio of acetic acid : NaOH= 0.0135:0.001568=8.61:1

this suggests excess is acetic acid and all the NaOH remains. moles of acetic acid reamining after reaction = 0.0135-0.01568= 0.011932 moles ,moles of sodium acetate formed = 0.001568

volume of solution after mixing = 18+1.6= 19.6 ml= 19.6/1000L= 0.0196L

concentrations : acetic acid ( designated as HA)= 0.011932/0.0196L=0.61M and A-( conjugate base of acetic acid ) coming from sodium acetate = 0.001568/0.0196=0.08M

Hendersen-Hasselbach equation can be written as

pH= pKa+ log {[A-]/[HA]}, Ka= 1.8*10-5, pKa=-log(Ka)= 4.74

pH= 4.74+log(0.08/0.61)=3.86

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