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Chapter 6: Q34P (page 143)

Write the autoprotolysis reaction of $H_{2} {SO}_{4}$ .

Short Answer

Expert verified

The autoprotolysis reaction for $H_{2} {SO}_{4}$ is:

role="math" localid="1663387364297" $2 H_{2} {SO}_{4} ⇌ {HSO}_{4}^{-} + H_{3} {SO}_{4}^{+}$

Step by step solution

01

Definition of Autoprolysis Reaction

Auto protolysis is the process in which a proton is transported between two indistinguishable molecules. One molecule acts as a Bronsted -Lowry base, and another acts as a Bronsted Lowry acid.

02

Autoprolysis Reaction

The autoprotolysis reaction for $H_{2} {SO}_{4}$ is:

$2 H_{2} {SO}_{4} ⇌ {HSO}_{4}^{-} + H_{3} {SO}_{4}^{+}$

In this reaction, two molecules of Sulfuric acid are reacted. One molecule transfers its proton to another molecule.

The proton donating molecule is said to be a Bronsted-Lowry acid, and the proton accepting molecule is a Bronsted-Lowry base.

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Most popular questions from this chapter

For the reaction ${Li}^{+} + H_{2} O ⇌ Li (OH) (aq) + H^{+}, K_{L} i = 2.3 \times 10^{- 14}$ . Combine this reaction with the reaction to find the equilibrium constant for the reaction ${Li}^{+} + {OH}^{-} ⇌ Li (OH) (aq) .$

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

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.

Calculate $[H^{+}]$ andpHfor the following solutions:

(a) $0.010 M {HNO}_{3}$

(b) $0.035 M KOH$

(c) $0.030 M HCI$

(d) $3.0 M HCI$

(e) $0.010 M [{({CH}_{3})}_{4} N^{+}] {OH}^{-}$

Tetramethylammonium hydroxide

See all solutions

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