Why is starch added near the endpoint rather than at the beginning?

Answer: Starch-iodine complex is reversible only when [I2] is low; adding starch early can cause it to bind strongly and slowly release I2, giving an indistinct endpoint At high [I2], starch forms a tightly bound, slow-to-release complex; adding it when titration is nearly complete gives a sharp blue-colorless endpoint.

The dashed line marks the equivalence point of a strong-acid/strong-base titration. The pH at that point is approximately:

Answer: 7 Strong acid + strong base produces a neutral salt and water; at the equivalence point pH = 7.

Titration of 25.0 mL of HCl with 1.0 M NaOH requires 30.0 mL to reach the equivalence point. The HCl concentration is

Answer: 1.2 M Ma Va = Mb Vb: Ma × 25.0 = 1.0 × 30.0 → Ma = 1.2 M.

AP Chemistry · Topic 4.6

Introduction to Titration Practice

Part of Chemical Reactions.(SPQ-4.B)

Practice questions

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Sample questions

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Sample 1difficulty 2/5
$A student determines the % NaOCl in a bleach by reacting 5.00 mL of bleach (density 1.05 g/mL) with excess KI in acidic solution. The liberated I2 is titrated to a starch endpoint (blue->colorless) with 0.1000 M Na2S2O3, requiring 26.45 mL. Reactions: OCl- + 2 I- + 2 H+ -> Cl- + I2 + H2O; I2 + 2 S2O3^2- -> 2 I- + S4O6^2-.$
$Why is starch added near the endpoint rather than at the beginning?$
- A
  $Starch turns colorless in the presence of OCl-$
- B
  $Starch reacts with thiosulfate$
- C
  $Starch is destroyed by I2 if added too soon$
- D
  $Starch-iodine complex is reversible only when [I2] is low; adding starch early can cause it to bind strongly and slowly release I2, giving an indistinct endpoint$
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Why
$At high [I2], starch forms a tightly bound, slow-to-release complex; adding it when titration is nearly complete gives a sharp blue->colorless endpoint.$
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Sample 2difficulty 3/5
$The dashed line marks the equivalence point of a strong-acid/strong-base titration. The pH at that point is approximately:$
- A
  $10$
- B
  $7$
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- C
  $4$
- D
  $1$
Why
$Strong acid + strong base produces a neutral salt and water; at the equivalence point pH = 7.$
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Sample 3difficulty 3/5
$Titration of 25.0 mL of HCl with 1.0 M NaOH requires 30.0 mL to reach the equivalence point. The HCl concentration is$
- A
  $0.5 M$
- B
  $1.0 M$
- C
  $0.83 M$
- D
  $1.2 M$
  check_circle
Why
$M_a V_a = M_b V_b: M_a \times 25.0 = 1.0 \times 30.0 \to M_a = 1.2 M.$
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Sample 4difficulty 4/5
$_2$
- A
  1. mL
- B
  $50.0 mL$
- C
  $12.5 mL$
- D
  $n_{H^+} = n_{OH^-}$
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Why
$^+$
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