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Chapter 6: Q3TY (page 132)

Use Table 6-1 to find the $[H^{+}]$ in water at $100 掳 C$ and at $0 掳 C$ .

Short Answer

Expert verified

$[H^{+}]$ in the water at $100 掳 C$ and at $0 掳 C$ are found to be $7.4 脳 10^{- 7}$ and , $3.4 脳 10^{- 8}$ respectively.

Step by step solution

01

Define the formula for finding Kb.

The value of can be calculated as:

$K_{b} = \frac{[{BH}^{+}] [OH]}{[B]}$

Where $K_{b}$ is called the base dissociation constant.

02

Step 2: Calculate the Kw.

Given:

At $100^{掳} C, K_{w} = 5.43 脳 10^{- 13}$

At $0^{掳} C K_{w} = 1.15 脳 10^{- 15}$

$K_{w} = [H^{+}] [O H^{-}]$

$[H^{+}]$ in the water at $100 掳 C$

$K_{w} = 5.43 脳 10^{- 14} = [H^{+}] [O H^{-}] = [x] [x] = [x]^{2} x = 5.43 脳 10^{- 14} = 7.4 脳 10^{- 7}$

At $0 掳 C$

$K_{w} = 1.15 脳 10^{- 15} = [H^{+}] [O H^{-}] = [x] [x] = [x]^{2} x = \sqrt{1.15 脳 10^{- 15}} = 3.4 脳 10^{- 8}$

Using temperature dependent of $K_{w}^{a}$ values $[H^{+}]$ in the water.

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Most popular questions from this chapter

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 $K = 1 脳 10^{11}$ , 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, $x 鈮� 0.00500 M$ . 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 ${BrO}_{3}^{-}$ .

Find $[{Cu}^{2 +}]$ in a solution saturated with ${Cu}_{4} {(OH)}_{6} ({so}_{4})$ if $[{OH}^{-}]$ is fixed at $1.0 脳 10^{- 6} M$ . Note that ${Cu}_{4} {(OH)}_{6} ({so}_{4})$ gives 1mol of ${SO}_{4}^{2 -}$ 4 mol for of ${Cu}^{2 +}$ .

${Cu}_{4} {(OH)}_{6} ({so}_{4}) (s) 鈬� 4 C u^{2 +} + 6 O H^{-} + S O_{4}^{2 -} K_{s p} = 2.3 脳 10^{- 69}$

Write the $K_{b}$ reaction of ${CN}^{-}$ Given that the $K_{a}$ value for $HCN$ is $6.2 脳 10^{- 10}$ calculate $K_{b}$ for ${CN}^{-}$ .

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

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

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

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

Don鈥檛 panic! Reverse the first reaction, add all the reactions

together, and see what cancels.

A solution contains $0.0500 {MCa}^{2 +}$ and $0.0300 {MAg}^{+}$ . Can 99%of ${Ca}^{2 +}$ be precipitated by sulphate without precipitating ${Ag}^{+}$ ? What will be the concentration of ${Ca}^{2 +}$ when ${Ag}_{2} {SO}_{4}$ begins to precipitate?

See all solutions

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