MasteringChemistry: Homework 2 httpsal session.masteringchemistry.com on.masteri

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MasteringChemistry: Homework 2 httpsal session.masteringchemistry.com on.masteringchemistry.com/myct/itemView?assignmentProblemiD-45543837 Cer Updates Available MCAT NSCS Eventbrite Do you want to restart to i these updates now or try t RIT MAIL Disney ESPN Humping Rei as Sweater ney ESPN Humping Rei as Sweater Apple Bing A Solution is Made By Mixing 9.00 Mmol (malli Masteri CHMB402 Spring 2015 Signed in as Lauren Jones Hel Homework2Titration of Weak Acid with Strong Base Resources previous 1 6 of 9 next Titration of Weak Acid with Strong Base Part A A certain weak acid, HA, with a K, value of 5.61 × 10-6, is titrated with NaOH. A solution is made by mixing 9.00mmol (millimoles) of HA and 3.00mmol of the strong base What is the resulting pH? Express the pH numerically to two decimal places. pH a Hints My Answers Give Up Review Part Part B More strong base is added until the equivalence point is reached. What is the pH of this solution at the equivalence point if the total volume is 48.OmL. Express the pH numerically to two decimal places. 10

Explanation / Answer

The addition of strong base to the weak acid will quantitatively form A-, the conjugate base. So, after the addition, you will have 6 mmol HA and 3 mmol A-. Then,

Ka = 5.61X10^-6 = [H+][A-]/[HA]
5.61X10^-6 = [H+]3.00/9.00
[H+] = 1.68X10^-5
pH = - log [H+] = 4.77

B) At the equivalence point you have a solution containing 8 mmol A- in a volume of 49 mL. A- reacts as a base as:

A- + H2O <---> HA + OH-
Kb = [HA][OH-]/[A-] = 1X10^-14 / 5.61X10^-6 = 1.78X10^-9
Let [OH-] = [HA] = x
[A-] = 0.008 mol / 0.049 L = 0.16 M

Then,
1.78X10^-9 = x^2 / 0.16
x = 1.69X10^-5 M = [OH-]
pOH = - log[OH-] = 4.77
pH = 14.00 - pOH = 9.23

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