The half-equivalence point of a titration occurs half way to the equivalence point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.480 moles of a monoprotic weak acid (K₁ = 6.7 x 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?

The half-equivalence point of a titration occurs half way to the equivalence point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. If 0.480 moles of a monoprotic weak acid (K₁ = 6.7 x 10-5) is titrated with NaOH, what is the pH of the solution at the half-equivalence point?

Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Concept explainers

Acid Base Titration

An acid-base titration is the method of quantitative analysis for determining the concentration of an acid and base by exactly neutralizing it with a standard solution of base or acid having known concentration.

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The half-equivalence point of a titration occurs half way to the equivalence point, where half of the analyte has reacted to form
its conjugate, and the other half still remains unreacted.
If 0.480 moles of a monoprotic weak acid (K₁ = 6.7 x 10-5) is titrated with NaOH, what is the pH of the solution at the
half-equivalence point?
pH =
4.89
Incorrect

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The equivalence point is defined as moles acid = moles base. So first, let's calculate the initial moles of base we used. molpase Mpase = Volsoiution Molarity base = 0.100 (M) Volume of solution= _50.00 mL Moles of base = _5.00 mmol So, we need 5.00 mmol of acid (titrant) to be delivered to the analyte. Since we know the molarity of the acid and moles of acid, we just need to solve for the volume of acid (titrant) to be added. molacid Voladded Macid Molarity acid = . 0.200 (M) Moles of acid = _5.00 (mmol) Volume of acid needed to reach equivalence point (mL)
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b) A solution was prepared by dissolving 1.5 g of Acetic acid (CH,COOH) in a 250cm solution. 50cm of this solution was allowed to react with 25cm of 0.1 NaOH solution (C-12, H-1, O-16) i) Determine the concentration of the acetic acid solution in moldm ii) iii) Write down an equation for the reaction. Calculate the number of moles of CH,COOH left unreacted and the number of moles of CH,COONA formed. iv) What name is given to the solution formed when all the NaOH has reacted? Calculate the Pp" of the solution in (iv) above (ka for CH,COOH = 1.8 x 10)