Chapter 6: QK-E (page 141)

Question: (a) From K_w in Table 6-1, calculate the pH of pure water at 0⁰,20⁰, and 40⁰C.
(b) For the reaction , at 25⁰C. In this equation, D stands for deuterium, which is the isotope ²H. What is the pD (=-log[D⁺]) for neutral D₂O?

Short Answer

Expert verified

(a) The pH of pure water at 0⁰,20⁰, and 40⁰C will be 7.469, 7.082,6.770 respectively.

(b) pD for neutral D₂O will be 7.435

Step by step solution

Determine the pH

$\begin{array}{rcl} [H^{+}] [O H^{-}] & = & K_{w} \\ L e t, [H^{+}] a n d [O H^{-}] & = & x \\ x . x & = & x^{2} \\ x^{2} & = & K_{w} \\ x & = & \sqrt{K_{w}} \end{array}$

From table 6.1 we get the values of K_w at different temperature

At 0⁰C K_w=1.15×10^-15

At 20⁰C K_w=6.88×10^-15

At 40⁰C K_w=2.88×10^-14

$\begin{array}{rcl} p H & = & - \log_{10} [H^{+}] \\ p H & = & - \log_{10} [x] \\ = & - \log_{10} [\sqrt{K_{w}}] \end{array}$

Therefore, the value of pH at 0⁰C will be

$p H = - \log_{10} [\sqrt{1.15 \times 10^{- 15}}] = 7.469 p H = - \log_{10} [\sqrt{6.88 \times 10^{- 15}}] = 7.082 p H = - \log_{10} [\sqrt{2.88 \times 10^{- 14}}] = 6.770$

pD for neutral D2O

In pure D₂O $[D^{+}] = [O D^{-}]$

Given $K = [D^{+}] [O D^{-}] = 1.35 \times 10^{- 15}$

$\begin{array}{rcl} K & = & [D^{+}] [O D^{-}] \\ K & = & 1.35 \times 10^{- 15} \\ [D^{+}] [O D^{-}] & = & 1.35 \times 10^{- 15} \\ [D^{+}] [D^{+}] & = & 1.35 \times 10^{- 15} \\ [D^{+}] & = & \sqrt{1.35 \times 10^{- 15}} \\ [D^{+}] & = & 3.67 \times 10^{- 8} M \\ p D & = & \log_{10} [D^{+}] \\ p D & = & 7.435 \end{array}$

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Question: (a) From K_w in Table 6-1, calculate the pH of pure water at 0⁰,20⁰, and 40⁰C.
(b) For the reaction , at 25⁰C. In this equation, D stands for deuterium, which is the isotope ²H. What is the pD (=-log[D⁺]) for neutral D₂O?

Short Answer

Step by step solution

Determine the pH

pD for neutral D2O

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Question: (a) From Kw in Table 6-1, calculate the pH of pure water at 00,200, and 400C.(b) For the reaction , at 250C. In this equation, D stands for deuterium, which is the isotope 2H. What is the pD (=-log[D+]) for neutral D2O?

Short Answer

Step by step solution

Determine the pH

pD for neutral D2O

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Physical Chemistry

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Question: (a) From K_w in Table 6-1, calculate the pH of pure water at 0⁰,20⁰, and 40⁰C.
(b) For the reaction , at 25⁰C. In this equation, D stands for deuterium, which is the isotope ²H. What is the pD (=-log[D⁺]) for neutral D₂O?