CHEM 14BL Lecture Notes - Lecture 3: Equivalence Point, Sodium Chloride, Titration

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redclam353

28 Feb 2017

School

UCLA

Department

Chemistry and Biochemistry

Course

CHEM 14BL

Professor

Russell Arlene

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Document Summary

Molarity: moles/l, independent of reaction, must know what you have, molecular weight = sum of atomic weights in molecule, is theoretical. Normality: equivalents/l, depends on reaction, can determine for an unknown, equivalent weight = weight of compound that reacts with 1 mole of h + and oh - , is experimental. For acids or bases with only one h + or oh - or that react with only one. For example, hcl, hno 3 naoh, nh 3 , and the equivalent weight must be equal to the molecular weight. For reacting with more than one mole of h + or oh - , the normality > molarity. Examples are h 3 po 4 , h 2 so 4 , and the equivalent weight is less than the molecular weight. You should write acid-base reactions for each of these compounds and determine the molecular weight and the equivalent weight.

Related Questions

1. The accepted dissociation constant for acetic acid is 1.8 x 10-5. One student's graphical determination of the Ka of this acid is 1.4 x 10-5. Calculate the percent error in this student's determination.

2.The equations used in this experiment would have to be modified slightly if an acid with a Ka of 1.0 x 10-3 or greater were used as an unknown. Assuming such a situation, explain the problems that would result from the use of the equations in their present form. What modifications would you make to give accurate results when using such an unknown?

a. The following two reactions happen in the solution. OH- + HAn LaTeX: \longrightarrow An- + H2O (Reaction 1) HAn LaTeX: \longrightarrow An- + H+ (Reaction 2) Within this lab, the acid used is a very weak acid with a Ka much smaller than 1 x 10-3. So the dissociation of the acid (Reaction 2) is negligible. The moles of An- at equilibrium is calculated as equal to the initial moles of OH-, and mole of HAn at equilibrium is calculated as the initial moles of HAn minus the initial moles of OH-.

b. When the Ka of the weak acid is equal or bigger than 1 x 10-3, the dissociation of acid (Reaction 2) is no longer negligible. In such case, An- is produced from both Reaction 1 and Reaction 2. The moles of An- produced from Reaction 1 is equal to the initial moles of OH-; and the moles of An- produced from Reaction 2 is equal to the moles of H+ (or H3O+), which can be calculated as molarity of H+ times liters of the solution mixture. So the moles of An- at equilibrium should be calculated as: moles of An- at equilibrium = initial moles of OH- + moles of H+ ; (Equation 1) And the moles of HA at equilibrium should be calculated as: moles of HA at equilibrium= initial moles of HA -moles of An- at equilibrium (Equation 2)

c. If we keep ignoring Reaction 2 even though the Ka of the weak acid is equal or bigger than 1 x 10-3, we will end up with an incorrectly smaller mole and smaller equilibrium concentration of An- and incorrectly bigger mole and bigger equilibrium concentration of HAn , thus an incorrectly smaller Ka. The problem can be avoided and an accurate determination of Ka can be obtained by calculating moles of An- at equilibrium and that of HAn by Equation 1 and Equation 2, respectively.

d. The following two reactions happen in the solution. OH- + HAn LaTeX: \longrightarrow An- + H2O (Equation 1) HAn LaTeX: \longrightarrow An- + H+ (Equation 2) Within this lab, the acid used is a very weak acid with a Ka much smaller than 1 x 10-3. So the dissociation of the acid (Equation 2) is negligible. The moles of An- at equilibrium is calculated as equal to the initial moles of OH-, and mole of HAn at equilibrium is calculated as the initial moles of HAn minus the initial moles of OH-. When the Ka of the weak acid is equal or bigger than 1 x 10-3, the dissociation of acid (Equation 2) is no longer negligible. In such case, An- is produced from both Equation 1 and Equation 2. The moles of An- produced from Equation 1 is equal to the initial moles of OH-; and the moles of An- produced from Equation 2 is equal to the moles of H+ (or H3O+), which can be calculated as molarity of H+ times liters of the solution mixture. So the moles of An- at equilibrium should be calculated as: moles of An- at equilibrium = initial moles of OH- + moles of H+ ; (Equation 3) And the moles of HA at equilibrium should be calculated as: moles of HA at equilibrium= initial moles of HA -moles of An- at equilibrium (Equation 4) If we keep ignoring Equation 2, we will end up with an incorrectly smaller equilibrium concentration of An- and incorrectly big equilibrium concentration of HAn , thus an incorrectly smaller Ka. The problem can be avoided and an accurate determination of Ka can be obtained by calculating moles of An- at equilibrium and that of HAn by Equation 3 and Equation 4 respectively.

e. all are correct

please help calculate percent yield

data:
step 2= .0906g
step 3= .101g
step 5= .1986g
step 7= .2105g
final product weight is= .0971g

I've already subtracted the weight of the vials so that only the weight of the compounds are shown

1. Determine the amount of solute in the following samples:

500 ml of 0.2 M KI (potassium iodide)

220 ml of 0.03 M H2SO4 (sulfuric acid)

2.5 l of 0.1 M MgNO3 (magnesium nitrate)

1 ml of 5 M HCl (hydrochloric acid)

This problem requires that you know what molarity (M) is. M (molarity) = # of moles of solutes / Liter of solution. You are given the volume. You should be able to obtain moles of solutes. Your answers can be in moles or grams (either one will be acceptable).

2. How many ml of 0.1 M HCl would be required to titrate 35 ml of 1 M NaOH to the end point?

This is an acid/base reaction.

1) First, write out the balanced chemical reaction for HCl + NaOH.

2) Second, figure out the number of moles of NaOH in 35 ml of 1 M NaOH.

3) How many moles of HCl is required for completely reacting with the moles of NaOH in Part 2).

3. What is the pH of a 0.03 M solution of the strong base NaOH?

1) This question requires that you know pH = - log [H₃O⁺], or simply as pH = - log [H⁺] (to make it easier).

2)So, you have to know how to use your scientific calculator to do log function (check Youtube).

3) This question also requires that you know: pOH = - log [OH^-].

4) This question also requires that you know: [H₃O⁺][OH^-] = 1.0 x 10^-14 or pH + pOH = 14

5) First step, determine the concentration of OH^- in 0.03M NaOH when it dissociates or breaks up into Na⁺ and OH^-.

6) This will give you [OH^-]. You can plug this into pOH = - log [OH^-] to get pOH.

7) Once you get pOH, then use (pH + pOH = 14) to get pH.

4. Balance the chemical reactions shown below and then write equilibrium expressions for each. All reactants and all products are in the gas phase.

CO + F₂ ------> COF₂

CH₄ + O₂ ------> CO₂ + H₂O

N₂O₄ -----> Br₂ + NO

BrNO -----> Br₂ + NO

1)This problem requires that you know how to balance chemical equations.

2)After you balance the equation, you need to know the equilibrium expression

5. Determine the oxidation states of each element comprising the following ionic compounds:

KClO₄

NaMnO₄

Sr(NO₃)₂

NaHCO₃

CH₄

CHEM 14BL Lecture Notes - Lecture 3: Equivalence Point, Sodium Chloride, Titration

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