Third Test (Midterm) March 2018 .- Fill the following table: Name Acid/Base/Neut

Question

Third Test (Midterm) March 2018 .- Fill the following table: Name Acid/Base/Neutral 0.000398 1 10 2.1 4.21 2.- In the following reaction identify the acid and its conjugate base and the base and its conjugate acid: te Base Acid Base 3.- Determine the hydrogen and hydroxide concentration, when we dissolve 0.50 moles of the AH in Conjugate acid 250 mL of solution. (Hint consider that [H+1 is small) pKa a. 6.5 425 c, 5.6 x H] a. 2.12 10* b. 1.06 102 c. 3.86 10s [OH-]- a. 1.08 * 1010 b. 8.82 . 10. c. 9.43 . 10 4.- if the measure pH of another solution of the previous buffer is 5.0 determine the value of the [A-]/[AH] (Hint: pH pKa log (A-]/ IAHI) Solution a. 3.55 b. 5.62 c. 2.85 d.4.88 5- The curvec 10.00 8.00 2.00 0.00 5.00 10.00 15.00 20.00 25.00 30 Volume et NaOH (mL.) a.- Determine the 2 values of the pKa pKal a. 1.0 b-3.0 c-4.00 pKa2 a.4.0 b.10.0 c. 6.0 b- Determine the pH and NaOH volume for both neutralizations points First neutralization ph- a.6 b.4.0 c. 6.0 Volume a. 21 b.35 c. 15 Second neutralization pH-a. 12 b. 6 c. 8.8 Volume a. 42 b-70 c. 35

Explanation / Answer

1. Table filling

formulas to use

pH = -log[H+]

pOH = -log[OH-]

pH + pOH = 14

when,

[H+] = 0.00398 M

pH = -log(0.00398) = 2.40

pOH = 14 - 2.40 = 11.60

[OH-] = 2.51 x 10^-12

acidic

when,

[OH-] = 1 x 10^-2 M

pOH = -log(1 x 10^-2) = 2

pH = 14 - 2 = 12

[H+] = 1 x 10^-12

basic

when,

pH = 2.1

[H+] = 7.94 x 10^-3 M

pOH = 14 - 2.1 = 11.9

[OH-] = 1.26 x 10^-12

acidic

when,

pOH = 4.21

[OH-] = 6.16 x 10^-5 M

pH = 9.79

[H+] = 1.62 x 10^-10 M

basic

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2. For the reaction,

CH3COOH + (CH3)2NH --> CH3COO-      +     (CH3)2NH2+

    acid                base        conjugate base        conjugate acid

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3. Ka = 5.623 x 10^-5

[AH] = 0.5 moles/0.250 L = 2 M

AH <==> A- + H+

let x amount AH dissociated

5.623 x 10^-5 = x^2/2

x = [H+] = 1.06 x 10^-2 M (b.)

[OH-] = 1 x 10^-14/1.06 x 10^-2 = 9.43 x 10^-13 M (c.)

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4. pKa = 4.25

5.0 = 4.25 + log([A-]/[HA])

([A-]/[HA]) = b. 5.62

5. For the plot

a. pKa1: b. 3.0 pKa2 : c. 6.0

b. first neutralization pH: b. 4 volume a. 21

second neutralization pH: c. 8.8 volume a. 42

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