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Chapter 6: Q30P (page 143)

Why is the pH of distilled water usually <7? How can you prevent this from happening?

Short Answer

Expert verified

The pH of distilled water usually <7 due to reaction of dissolved carbon dioxide with water.

Step by step solution

01

Concept Used

ph is a measurement of the level of acid or alkali in a substance or liquid.

pH is the negative logarithm of the concentration of $H^{+}$ ions in solution.

To calculate the pH of an aqueous solution we need to know the concentration of the Hydronium ion in moles per liter (morality).

Therefore, the pH of a substance is given by:

pH = - log [H₃O⁺].

02

Explanation for pH of distilled water being less than 7.

Carbon dioxide $({CO}_{2})$ is present in the atmosphere. Thus, dissolved ${CO}_{2}$ lowers pH of the dissolved water by reacting with it to form carbonic acid. Hence, the concentration of hydronium ion decreases in a solution the pH of distilled water becomes less than 7.

03

Prevention of formation of carbon dioxide.

To prevent the formation of carbonic acid, water can be distilled under an inert atmosphere so that carbon dioxide is not dissolved in water. Most of the dissolvedcarbon dioxide can be eliminated by boiling the distilled water.

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Most popular questions from this chapter

Identify the conjugate acid-base pairs in the following reactions:

(a) $H_{3} {NCH}_{2} {CH}_{2} {NH}_{3} + H_{2} O ⇌ H_{3} {NCH}_{2} {CH}_{2} {NH}_{2} + H_{3} O^{+}$

(b) Benzoicacid+ Pyridine $⇌$ Benzoate+ Pyridinium

Write the formulas and names for three classes of weak acids and two classes of weak bases.

The equilibrium constant for the reaction of ${NH}_{3} (aq) + H_{2} O ⇌ {NH}_{4}^{+} + {OH}^{-}$ is $K_{b} = 1.479 \times 10^{- 5}$ at $5 ° C$ and $1.570 \times 10^{- 5}$ at $10 ° C$ .

(a) Assuming $∆ H °$ and $∆ S °$ are constant in the interval $5 ° - 10 ° C$ (probably a good assumption for small $∆ T$ ), use Equation 6-9 to find $∆ H °$ for the reaction in this temperature range.

(b) Describe how Equation 6-9 could be used to make a linear graph to determine $∆ H °$ , if $∆ H °$ and $∆ S °$ were constant over some temperature range

The planet Aragonose (which is made mostly of the mineral

aragonite, or ${CaCO}_{3}$ ) has an atmosphere containing methane and

carbon dioxide, each at a pressure of 0.10 bar. The oceans are

saturated with aragonite and have a concentration of $H_{1}$ equal to

$1.8 \times 10^{27} M$ . Given the following equilibria, calculate how many

grams of calcium are contained in 2.00 L of Aragonose seawater.

$\begin{matrix} {CaCO}_{3} (s, aragonite) & ⇌ {Ca}^{2 +} (aq) + {CO}_{3}^{2 -} (aq) & K_{sp} = 6.0 \times 10^{- 9} \\ {CO}_{2} (g) & ⇌ {CO}_{2} (aq) & K_{{CO}_{2}} {CO}_{2}) = 3.4 \times 10^{- 2} \\ {CO}_{2} (aq) + H_{2} O (l) & ⇌ {HCO}_{3}^{-} (aq) + H^{+} (aq) & K_{1} = 4.5 \times 10^{- 7} \\ {HCO}_{3}^{-} (aq) & ⇌ H^{+} (aq) + {CO}_{3}^{2 -} (aq) & K_{2} = 4.7 \times 10^{- 11} \end{matrix}$

Don’t panic! Reverse the first reaction, add all the reactions

together, and see what cancels.

Reaction 6-8 is allowed to come to equilibrium in a solution initially containing $0.0100 {MBrO}_{3}^{-}$ , $0.0100 {MCr}^{3 +}$ and 1.00MH⁺. To find the concentrations at equilibrium, we construct the table at the bottom of the page showing initial and final concentrations. We use the stoichiometry coefficients of the reaction to say that if $x m ol$ of ${Br}^{-}$ are created, then we must also make x mol of ${Cr}_{2} O_{7}^{2 -}$ and 8x mol of H⁺. To produce x mol of ${Br}^{-}$ , we must have consumed x mol of ${Br}^{-} O_{3}^{-}$ and 2x mol of Cr³⁺.

(a) Write the equilibrium constant expression that you would use to solve for x to find the concentrations at equilibrium. Do not try to solve the equation.

(b) Because $K = 1 \times 10^{11}$ , we suppose that the reaction will go nearly "to completion." That is, we expect both the concentration of ${Br}^{-}$ and ${Cr}_{2} O_{7}^{2 -}$ to be close to 0.00500M an equilibrium. (Why?) That is, $x \approx 0.00500 M$ . With this value of $x, [H^{+}] = 1.00 + 8 x = 1.04 M$ and $[{BrO}_{3}^{-}] = 0.0100 - x = 0.0050 M$ . However, we cannot say $[{Cr}^{3 +}] = 0.0100 - 2 x = 0$ , because there must be some small concentration of ${Cr}^{3 +}$ at equilibrium. Write $[{Cr}^{3 +}]$ for the concentration of ${Cr}^{3 +}$ and solve for $[{Cr}^{3 +}]$ . The limiting reagent in this example is ${Cr}^{3 +}$ . The reaction uses up before consuming ${BrO}_{3}^{-}$ .

See all solutions

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