15.4 Titrations and pH curves Flashcards

1
Q

What is a titration?

A

Its when we add a solution of known concentration (titrant) in a solution of unknown concentration to analyze it (analyte).

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2
Q

What is an equivalence point?

A

The point when enough titrant is added to react with the analyte (mol of titrant = mol of analyte)

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3
Q

What is a acid-base titration?

A

analyte is a base/acid and titrant is a strong base/strong acid

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4
Q
Consider the titrant of 50.0 mL of 0.200M HNO3 with 0.100M NaOH. Calculate the pH of the solution where specific volumes of 0.100M NaOH have been added.
point 1= no NaOH added
point 2= 10.0mL added
point 3=100.0 mL added(total)
point 4= 150mL added (total) R
A

Point 1=0.699
point 2=0.82
point 3=7
point 4= 12.40

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5
Q

Determine the concentration of HCI solution. 0.100M NaOH is added slowly into an HCL solution of unknown concentration and of 35 mL of volume. The equivalence point comes when we add 35 mL of titrant. (vertical line in the pH titration curve) O

A

Concentration HCl = 0.100M

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6
Q

50.00 ml of 0.0884M HNO3 with 0.0980M NaOH.
HNO3(aq) + NaOH(aq) → NaNO3 (aq) + H2O(liq)
At the Equivalence pt pH=7 experimentally. Calculate the equivalence pt volume in the titration of 50.00 ml of 0.0884M nitric acid with 0.0980M NaOH. O

A

Volume = 45.1* 10^-3 L

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7
Q

calculate the pH in the titration of 20.00ml of 0.125M HCl after the addition of: a) 0 b)2.00 c) 10.00 ml of 0.250M NaOH. Calculate using milimoles. O

A

a) 0.9
b) 1.04
c) 7

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8
Q

Draw a strong acid strong base titration curve. Indicate the eq. point and the pH

A

check book.

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9
Q

Draw a weak acid strong base titration curve. Indicate the eq. point, the buffer region and other things to know.

A

check book.

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10
Q

Write the buffer region equation.

A

pH = aka + log( [A-]/[HA] )

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11
Q

At half-equivalence point, [HA] = [A-} why? HA= weak acid A- = conj. base

A

The number of mols of the remaining weak acid is equal to the number of mols of the conjugate base formed.

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12
Q

calculate the pH during the titration of 25.0 ml of 0.50M formic acid
(Ka= 1.8*10-4) with 0.50M NaOH, calculate the pH after:
a) 0 b) 12.5 c) 25.0 ml of NaOH have been added O

A

a) 2.02
b) 3.74
c) 8.57

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13
Q

40.0 ml of 0.10M NH3 + 0.10M HCl ( a) 0. b)20. c)40. d)60 ml ). Calculate pH.

A

a) 11.12
b) 9.25
c) 5.28
d) 1.70

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14
Q

What is a monoprotic acid or base?

A

an acid or base with only one H or OH

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