9) An unknown weak acid is titrated to its equivalence point using 10 mL of 1M NaOH. At the equivalence point, the pHof the 100 mL solution is 9. What is the pKa of the acid? Knowing that, what was the pH of the original 90 mL solution?mol NaOH = (10ML) (IM) = 10 mmol[x-] = 10mmol100mL= 0.1Mpka:pH

In this question, we are given a titration scenario involving an unknown weak acid and NaOH. Here's…

Answered: 9) An unknown weak acid is titrated to…

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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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A 100 mL sample contains Na2CO3 and NaOH. A 25 mL aliquot was taken and phenolphthalein was added as indicator into the solution. The sample solution was titrated with 13.0 mL of 0.10 M HCl. After the disappearance of the pink color, methyl orange was added and the sample solution was titrated with 4.0 mL of 0.10 M HCl. Calculate the amounts of Na2CO3 and NaOH in the sample as mg/100 mL. (write all chemical reactions, Na:23, C:12, O:16, H:1 g/mol).
For the titration of 20.0 mL of 0.100 M carbonic acid H2CO3 (Ka1 = 4.3 ×10-7, Ka2 = 5.6× 10-11) with 0.125 M NaOH, calculate the: Initial pH: pH at first half- equivalence point: Volume of added base at the first equivalence point: pH at second half-equivalence point: Volume of added base at the second equivalence point:
In order to determine the Ka of a weak, monoprotic acid, a student measured the pH of a solution of this weak, monoprotic acid to be 3.617. Based on the full- neutralization titration portion of the experiment, it was found that the total molar concentration of this acid solution was 2.632 M. Using these measurements, what would be the Ka for this weak, monoprotic acid? Ka X 10-7 Please enter as your answer only the underlined numerical part of the Ka above. Round your answer to two decimal places.
A solution of 25.0 mL of 0.095 M hydroxylamonlum lon (weak acid) was titrated with 0.110 M NaOH. The Ka of hydroxylamonium Is 1.1 x 10-6. Calculate the following: What is the volume of NaOH solution at the equivalence point? 21.6 What is the pH after 10 mL of the NaOH is added? 10.85 What is the pH at the equivalence point? 9.85 What is the pH after 25 mL of the NaOH is added? 11.87 X X mL
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MISSED THIS? Read Section 18.4 (Page); Watch KCV 18.4B. Consider the curve shown here for the titration of a weak monoprotic acid with a strong base. Hd 14 12 10 8 6420 80 120 160 40 Volume of base added (mL) The figure shows pH as a function of volume of base added. The pH is measured from 0 to 14 on the y-axis, while the volume of base added is measured from 0 to 160 milliliters on the x-axis. The curve of the plot goes up gradually from pH 2.3 at 0 milliliters to pH 5.8 at 49 milliliters, next the plot goes up steeply to pH 11.8 at 51 milliliters, and finally the plot goes up gradually to pH 12.2 at 160 milliliters. ✓ Correct An equilibrium problem based on the initial concentration and Ka of a weak acid should be solved for aqueous solution of the weak acid. This is true for the initial state of titration, when no base has been addition of a small amount of a strong base turns this solution into a buffer whose pH is calculated in way. Part D At what volume of added base does pH =…
A The graph shows the titration curves of a strong acid and a weak acid. Estimate the pk, of the weak acid from its titration curve V 14 12 10 pH 12086420 Volume of strong base (ml) 1st attempt Write the estimated pK, to one decimal place 1 OF 8 QUESTIONS COMPLETED 82 W S # 3 Z X E 80 FI D $ 4 C R F K5 V 4 FS T G A 6 B S Y 27 & H 8 U N 8 00 DHI 14 M 9 K TV O hi See Periodic Table O L SUBMIT ANSWER P ** FIL
In the titration of a 25.0-mL sample of 0.145 M HCOOH with 0.122 M NaOH, where K, HCOOH = 1.8x10-4 %3D the volume of NAOH added at one-half the equivalence point is about 27.4 mL O the volume of NaOH added at one-half the equivalence point is about 7.45 mL O the volume of NaOH added at one-half the equivalence point is about 12.6 mL O the volume of NaOH added at one-half the equivalence point is about 14.9 mL
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If the pH at the 1/2 equivalence point is 4.19, what would be the Ka?
Suppose we titrate 20.0 mL of 0.100 M HCHO2 (Ka = 1.8 x 10-4) with 0.100 M NaOH. Calculate the pH. Before any base is added: pH = When half of the HCHO₂ has been neutralized: pH = After a total of 15.0 mL of base has been added: pH At the equivalence point: pH =

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