Question

1. complete the reaction \\( \ce{h2so4 + 2 h2o -> 2 h3o^{+1} + \\_\\_1\\_\\_} \\) assume there is complete ionization of \\( \ce{hso4^{-1}} \\)\
2. what is the ph of a 0.001000 m \\( \ce{h2so4} \\) solution ?\

first, find the \\( \ce{h3o^{+1}} \\) using stoichiometry\
\\( \\_\\_2\\_\\_ \\) mol \\( \ce{h3o^{+1}} \\)\
\\( (1.0000 \times 10^{-3} \text{ m } \ce{h2so4})(-------------) = \\_\\_3\\_\\_ \text{ m } \ce{h3o^{+1}} \\)\
\\( \\_\\_4\\_\\_ \\) mol \\( \ce{h2so4} \\) \\( \text{ph} = -\log(\\_\\_5\\_\\_) = \\_\\_6\\_\\_ \\)\
a. acetic \quad b. hydrochloric \quad c. phosphoric \quad d. perchloric \quad e. hydrofluoric\
f. sulfuric \quad g. sulfurous \quad h. hydrosulfuric \quad i. \\( \ce{oh^{-1}} \\) \quad j. \\( \ce{h3o^{+1}} \\) \quad k. \\( \ce{so4^{-2}} \\)\
l. \\( \ce{mg^{+2}} \\) \quad m. \\( \ce{cl^{-1}} \\) \quad n. \\( \ce{hpo3^{-2}} \\) \quad o. \\( \ce{h2po3^{-1}} \\) \quad p. \\( \ce{co3^{-2}} \\) \quad q. \\( \ce{hco3^{-1}} \\)\
r. \\( \ce{h2s} \\) \quad s. \\( \ce{hs^{-1}} \\) \quad t. \\( \ce{s^{-2}} \\) \quad u. 0.2500 \quad v. 0.1000 \quad w. 0.02500\
x. \\( 1.000 \times 10^{-3} \\) \quad y. 3.0000 \quad z. \\( 1 \times 10^{-14} \\) \quad aa. 2.6990 \quad bb. \\( 1.000 \times 10^{-11} \\)\
cc. 11.0000 \quad dd. 2 \quad ee. 1 \quad ff. \\( 2.000 \times 10^{-3} \\) \quad gg. \\( 5.000 \times 10^{-12} \\)\
hh. 11.3010 \quad ii. 0.01250 \quad jj. 0.22100 \quad kk. 0.05656 \quad ll. acid\
mm. conjugate acid \quad oo. base \quad pp. conjugate base

Explanation:

Step1: Complete the reaction

According to the law of conservation of mass and charge, for the reaction \(H_{2}SO_{4}+2H_{2}O
ightarrow2H_{3}O^{+1}+\)_1_, the missing ion is \(SO_{4}^{-2}\) (since there are 2 \(H^{+}\) from \(H_{2}SO_{4}\) combining with 2 \(H_{2}O\) to form 2 \(H_{3}O^{+}\), and the remaining part is \(SO_{4}^{2 -}\)).

Step2: Find the \(H_{3}O^{+}\) concentration

From the reaction \(H_{2}SO_{4}+2H_{2}O
ightarrow2H_{3}O^{+}+SO_{4}^{2 -}\), the mole ratio of \(H_{2}SO_{4}\) to \(H_{3}O^{+}\) is \(1:2\).
For a \(0.001000M\ H_{2}SO_{4}\) solution, \([H_{3}O^{+}]=(1.000\times 10^{- 3}M\ H_{2}SO_{4})\times\frac{2\ mol\ H_{3}O^{+}}{1\ mol\ H_{2}SO_{4}} = 2.000\times 10^{-3}M\)

Step3: Calculate the pH

The formula for pH is \(pH=-\log[H_{3}O^{+}]\). Substituting \([H_{3}O^{+}]=2.000\times 10^{-3}M\) into the formula, we get \(pH =-\log(2.000\times 10^{-3})=2.6990\)

Answer:

1. K. \(SO_{4}^{-2}\)
2. 2. \(2\)
3. FF. \(2.000\times 10^{-3}\)
4. EE. \(1\)
5. FF. \(2.000\times 10^{-3}\)
6. AA. \(2.6990\)