Chapter 11: Q24P (page 260)

A 100.0-mL aliquot of 0.100 M diprotic acid H₂A (pK₁ = 4.00, pK₂ = 8.00) was titrated with 1.00 M NaOH. Find the pH at the following volumes of base added and make a graph of pH versus Vb: Vb = 0, 1, 5, 9, 10, 11, 15, 19, 20, and 22 mL.

Short Answer

Expert verified

The pH of the solution is tabulated below

Va	0	1	5	9	10	11	15	19	20	22
pH	2.51	3.05	4.00	4.95	6.00	7.05	8.00	8.95	10.46	12.21

The graph is shown below

Step by step solution

Determine initial concentration

A 100.0-mL aliquot of 0.100 M diprotic acid H₂A (pK₁ = 4.00, pK₂ = 8.00) was titrated with 1.00 M NaOH

$\begin{array}{rcl} K_{1} & = & \frac{[H^{+}] [H A^{-}]}{[H_{2} A]} \\ K_{1} & = & 10^{- 4} \\ \frac{[H^{+}] [H A^{-}]}{[H_{2} A]} & = & 10^{- 4} \\ \frac{x^{2}}{0.1 - x} & = & 10^{- 4} \\ x & = & 3.11 \times 10^{- 3} M \end{array}$

Determine the pH

Addition of 0 mL base

$\begin{array}{rcl} [H^{+}] & = & 3.11 \times 10^{- 3} M \\ p H & = & 2.51 \end{array}$

Addition of 1 mL base

$p H = p K_{1} + \log \frac{[H A^{-}]}{[H_{2} A]} = 4.00 + \log \frac{1}{9} = 3.05$

Addition of 10 mL base

Concentration of HA^- will be

$\begin{array}{rcl} \frac{100}{110} (0.1) & = & 0.0909 M \\ [H^{+}] & = & \sqrt{\frac{10^{- 4} . 10^{- 8} (0.0909) + 10^{- 4} . 10^{- 14}}{10^{- 6} + 0.0909}} \\ = & 9.99 \times 10^{- 7} \\ p H & = & 6.00 \end{array}$

Addition of 11 mL base

$\begin{array}{rcl} p H & = & p K_{2} + \log \frac{[A^{2 -}]}{[H A^{-}]} \\ = & 8.00 + \log \frac{1}{9} \\ = & 7.05 \end{array}$

Addition of 20 mL base

$\begin{array}{rcl} \frac{x . x}{0.0833 - x} & = & \frac{K_{w}}{K_{2}} \\ x & = & 2.88 \times 10^{- 4} M \\ p H & = & - \log \frac{K_{w}}{x} \\ = & 10.46 \end{array}$

Addition of 22 mL base

$\begin{array}{rcl} [O H^{-}] & = & (\frac{2}{122}) (1) = 1.64 \times 10^{- 2} M \\ p H & = & 12.21 \end{array}$

Following the above calculations, the pH of the solution is tabulated below

Va	0	1	5	9	10	11	15	19	20	22
pH	2.51	3.05	4.00	4.95	6.00	7.05	8.00	8.95	10.46	12.21

The graph is shown below

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A 100.0-mL aliquot of 0.100 M diprotic acid H₂A (pK₁ = 4.00, pK₂ = 8.00) was titrated with 1.00 M NaOH. Find the pH at the following volumes of base added and make a graph of pH versus Vb: Vb = 0, 1, 5, 9, 10, 11, 15, 19, 20, and 22 mL.

Short Answer

Step by step solution

Determine initial concentration

Determine the pH

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A 100.0-mL aliquot of 0.100 M diprotic acid H2A (pK1 = 4.00, pK2 = 8.00) was titrated with 1.00 M NaOH. Find the pH at the following volumes of base added and make a graph of pH versus Vb: Vb = 0, 1, 5, 9, 10, 11, 15, 19, 20, and 22 mL.

Short Answer

Step by step solution

Determine initial concentration

Determine the pH

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Kinetics

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Ionic and Molecular Compounds

Study anywhere. Anytime. Across all devices.

A 100.0-mL aliquot of 0.100 M diprotic acid H₂A (pK₁ = 4.00, pK₂ = 8.00) was titrated with 1.00 M NaOH. Find the pH at the following volumes of base added and make a graph of pH versus Vb: Vb = 0, 1, 5, 9, 10, 11, 15, 19, 20, and 22 mL.