Reaction: HCl + NaOH → NaCl + H2OEnd pointpH: 8.00PHEquivalence point(using your graphs)pH: 8.5mLVolume of NaOH: 20mLVolume of NaOH: 22 mLDetermine the molarity of NaOH (end point):(show calculation)Determine the molarity of NaOH (equivalence point):(show calculation)What can you conclude about the end point and the equivalence point? Is there a titrationerror?

To determine the molarity of NaOH at both the end point and the equivalence point, we need to use…

Answered: Reaction: HCl + NaOH → NaCl + H2O End…

Chemistry

10th Edition

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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Write out a chemical equation. (The chemical equation, for example, is NaOH + HCl ® HOH + NaCl.) Explain what the equivalence point is, what the pH is at the equivalence point in this reaction and why the pH is what it is.
100.0 mL of HCI required 25.00 mL of 0.2000 M Mg(OH)2 to reach pH 7. What is the concentration of the HCI ? 2 HCI + Mg(OH)₂ MgCl₂ + 2 H₂O The neutralization equation is There is enough information to calculate the moles of Mg(OH)2 ( L Mg(OH)₂ (- concentration = Then use stoichiometry to convert moles of Mg(OH)₂ into moles of HCI mol HCI a. acid i. 0.2000 p. 0.01000 mol Mg(OH)₂ (- b. base c. HCI d. NaOH mol Mg(OH)2 We now can calculate the concentration of HCI since we have both the moles of HCI and the volume of HCI j. 1.000 k. 0.005000 q. 0.001000 1 L mol HCI LHCI r. 58.31 = mol Mg(OH)2 e. NaCl I. 100.0 f. H₂O m. 0.1000 s. 5.831 = g. 25.00 mol/L n. 0.05000 h. 0.02500 mol Mg(OH)2 o. 2.000 mol HCI
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Find the pOH of the following solutions: 5) A7.98 x 10-2 M solution of LiOH (lithium hydroxide). 6) A7.98 x 10-2 M solution of Ba(OH)2 (barium hydroxide). 7) 735 mL of a solution containing 0.34 moles of NaOH. 8) 1098 mL of a solution containing 0.2543 moles of hydrobromic acid. --- ----.
Q6
Help me answer this one question:
A diprotic acid is titrated with a strong base. The pH at the first half-equivalence point is (3.69x10^0). The pH at the second half-equivalence point is (8.9000x10^0). What is the value of ka2? Enter your answer using 3 significant figures.

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